Tag: Partial Fractions

Find the slope and area under the graph for a given function

by RoRi

December 17, 2015

Let

$f(x) = \sqrt{\frac{4x+2}{x(x+1)(x+2)}} \qquad \text{for} \quad x> 0.$

Determine the slope of the graph of $f$ at the point with $x$ -coordinate 1.
Find the volume of the solid of revolution formed by rotating the region between the graph of $f(x)$ and the interval $[1,4]$ about the $x$ -axis.

To take this derivative, using logarithmic differentiation will be easier,
$\begin{align*} \log (f(x)) &= \log \left( \sqrt{ \frac{4x+2}{x(x+1)(x+2)}} \right) \\[9pt] &= \frac{1}{2} \left( \log (4x+2) - \log (x(x+1)(x+2)) \right) \\[9pt] &= \frac{1}{2} \left( \log (4x+2) - \log x - \log (x+1) - \log (x+2) \right). \end{align*}$

Then differentiating both sides we have,

$\begin{align*} &&\frac{f'(x)}{f(x)} &= \frac{1}{2} \left( \frac{4}{4x+2} - \frac{1}{x} - \frac{1}{x+1} - \frac{1}{x+2} \right) \\[9pt] \implies && f'(x) &= \frac{1}{2} \sqrt{\frac{4x+2}{x(x+1)(x+2)}} \left( \frac{2}{2x+1} - \frac{1}{x} - \frac{1}{x+1} - \frac{1}{x+2} \right). \end{align*}$

So, to find the slope at the point with $x = 1$ we evaluate,

$f'(1) = \frac{1}{2} \left( \frac{2}{3} - 1 - \frac{1}{2} - \frac{1}{3} \right) = -\frac{7}{12}.$
First, the integral to compute the volume of the solid of revolution is,
$\begin{align*} V &= \pi \int_1^4 (f(x))^2 \, dx \\[9pt] &= \int_1^4 \frac{\pi(4x+2)}{x(x+1)(x+2)} \, dx. \end{align*}$

To evaluate this we use the partial fraction decomposition,

$\frac{2x+1}{x(x+1)(x+2)} = \frac{A}{x} + \frac{B}{x+1} + \frac{C}{x+2}.$

This gives us the equation

$A(x+1)(x+2) + B(x)(x+2) + C(x)(x+1) = 2x+1.$

Evaluating at $x = 0$ , $x = -1$ , and $x = -2$ we obtain

$A = \frac{1}{2}, \quad B = 1, \quad C = -\frac{3}{2}.$

Therefore, we have

$\begin{align*} V &= 2 \pi \int_1^4 \left( \frac{1}{2x} + \frac{1}{x+1} - \frac{3}{2(x+2)} \right) \, dx \\[9pt] &= 2 \pi \left( \frac{1}{2} \int_1^4 \frac{1}{x} \,dx + \int_1^4 \frac{1}{x+1} \, dx - \frac{3}{2} \int_1^4 \frac{1}{x+2} \, dx \right) \\[9pt] &= \pi \log x \Bigr \rvert_1^4 + 2 \pi \log |x+1| \Bigr \rvert_1^4 - 3 \pi \log |x+2| \Bigr \rvert_1^4 \\[9pt] &= \pi \log 4 + 2 \pi (\log 5 - \log 2) - 3 \pi (\log 6 - \log 3) \\[9pt] &= 2 \pi \log 2 + 2 \pi \log 5 - 2 \pi \log 2 - 3 \pi \log (2 \cdot 3) + 3 \pi \log 3 \\[9pt] &= 2 \pi \log 5 - 3 \pi \log 2 - 3 \pi \log 3 + 3 \pi \log 3 \\[9pt] &= 2 \pi \log 5 - 3 \pi \log 2 \\[9pt] & = \pi (\log 25 - \log 8) \\[9pt] & = \pi \log \frac{25}{8}. \end{align*}$

Evaluate the integral of (x² + x)^1/2 / x

by RoRi

December 17, 2015

Compute the following integral.

$\int \frac{\sqrt{x^2+x}}{x} \, dx.$

First, we have

$\int \frac{\sqrt{x^2+x}}{x} \, dx = \int \frac{\sqrt{x}\sqrt{x+1}}{x} \, dx = \int \sqrt{\frac{x+1}{x}} \, dx.$

Then, we make a substitution, letting

$\begin{align*} &&u &= \frac{x+1}{x} \\ \implies && du &= -\frac{1}{x^2} \, dx \\ \implies && dx &= \frac{1}{(1-u)^2}. \end{align*}$

This gives us

$\int \frac{\sqrt{x^2+x}}{x} \, dx = -\int \frac{\sqrt{u}}{(1-u)^2} \, du.$

Next, we make another substitution, letting $t = \sqrt{u}$ and $dt = \frac{1}{2\sqrt{u}} \, du$ . This gives us,

$\begin{align*} -\int \frac{\sqrt{u}}{(1-u)^2} \, du &= -2 \int \frac{t^2}{(1-t^2)^2} \, dt \\[9pt] &= -2 \int \frac{t^2}{(t^2-1)^2} \, dt \\[9pt] &= -2 \int \frac{t^2}{(t+1)^2 (t-1)^2} \, dt. \end{align*}$

We can then use partial fractions on the integrand,

$\frac{t^2}{(t+1)^2 (t-1)^2} = \frac{A}{t+1} + \frac{B}{(t+1)^2} + \frac{C}{t-1} + \frac{D}{(t-1)^2}.$

This gives us the equation

$A(t+1)(t-1)^2 + B(t-1)^2 + C(t+1)(t-1) + D(t+1)^2 = t^2.$

Solving this for $A, B, C$ , and $D$ we obtain

$A = -\frac{1}{4}, \quad B = C = D = \frac{1}{4}.$

Therefore,

$\begin{align*} -2 \int \frac{t^2}{(t^2-1)^2} \, dt &= -2 \int \left( \frac{-1}{4(t+1)} + \frac{1}{4(t+1)^2} + \frac{1}{4(t-1)} + \frac{1}{4(t-1)^2} \right) \, dt \\[9pt] &= -\frac{1}{2} \left( \int \frac{1}{(t-1)^2} \, dt + \int \frac{1}{t-1} \, dt + \int \frac{1}{(t+1)^2} \, dt - \int \frac{1}{t+1} \, dt \right) \\[9pt] &= \frac{1}{2(t-1)} - \frac{1}{2} \log |t-1| + \frac{1}{2(t+1)} + \frac{1}{2} \log|t+1| + C \\[9pt] &= \frac{1}{2} \left( \frac{1}{t-1} + \frac{1}{t+1} \right) + \frac{1}{2} \log \left| \frac{t+1}{t-1} \right| + C \\[9pt] &= \frac{1}{2} \left( \frac{2t}{t^2-1} \right) + \frac{1}{2} \log \left| \frac{\sqrt{u} + 1}{\sqrt{u} - 1} \right| + C \\[9pt] &= \frac{1}{2} \left( \frac{2\sqrt{u}}{u-1} \right) + \frac{1}{2} \log \left| \frac{\sqrt{\frac{x+1}{x}} + 1}{\sqrt{\frac{x+1}{x}} - 1} \right| + C\\[9pt] &= \sqrt{x^2+x} + \frac{1}{2} \log \left| \frac{\sqrt{x+1} + \sqrt{x}}{\sqrt{x+1} - \sqrt{x}} \right| \\[9pt] &= \sqrt{x^2+x} + \frac{1}{2} \log \left( \sqrt{x+1} + \sqrt{x} \right)^2 + C\\[9pt] &= \sqrt{x^2+x} + \frac{1}{2} \log \left( 2 \sqrt{x^2+x} + 2x + 1 \right) + C. \end{align*}$

Evaluate the integral of (3 – x²)^1/2 / x

by RoRi

December 17, 2015

Compute the following integral.

$\int \frac{\sqrt{3-x^2}}{x} \, dx.$

First, we multiply the numerator and denominator by $\sqrt{3-x^2}$ and do some rearranging to get a friendlier integral,

$\begin{align*} \int \frac{\sqrt{3-x^2}}{x} \, dx &= \int \frac{3-x^2}{x \sqrt{3-x^2}} \, dx \\[9pt] &= - \int \frac{x}{\sqrt{3-x^2}} \, dx + 3 \int \frac{1}{x \sqrt{3-x^2}} \, dx. \end{align*}$

For the integral on the left we make the substitution $u = 3-x^2$ , $du = -2x \, dx$ and obtain

$-\int \frac{x}{\sqrt{3-x^2}} \, dx = \frac{1}{2} \int \frac{du}{\sqrt{u}} = \sqrt{u} = \sqrt{3-x^2}.$

Then, for the integral on the right we make the substitution $u = \sqrt{3-x^2}$ , $du = \frac{-x}{\sqrt{3-x^2}} \, dx$ . This gives us

$x = \sqrt{3-u^2}, \qquad dx = -\frac{\sqrt{3-x^2}}{x}.$

Therefore for the integral on the right we have

$\begin{align*} 3 \int \frac{1}{x\sqrt{3-x^2}} \, dx &= 3 \int \frac{1}{3-u^2} \, du \\[9pt] &= -3 \int \frac{1}{(\sqrt{3} + u)(\sqrt{3} - u)} \, du. \end{align*}$

Now, we have to use partial fractions on the integrand,

$\frac{1}{(\sqrt{3} + u)(\sqrt{3} - u)} = \frac{A}{\sqrt{3} + u} + \frac{B}{\sqrt{3} - u}.$

This gives us the equation

$A(\sqrt{3} - u) + B(\sqrt{3} + u) = 1 \quad \implies \quad A = B = \frac{1}{2\sqrt{3}}.$

So,

$\begin{align*} -3 \int \frac{1}{(\sqrt{3} + u)(\sqrt{3} - u)} \, du &= -3 \int \left(\frac{1}{2 \sqrt{3} (\sqrt{3} + u)} - \frac{1}{2 \sqrt{3} (\sqrt{3} - u)} \right) \, du \\[9pt] &= -\frac{\sqrt{3}}{2} \left( \frac{1}{\sqrt{3} + u} \, du + \int \frac{1}{\sqrt{3} - u} \, du \right) \\[9pt] &= -\frac{\sqrt{3}}{2} \left( \log | \sqrt{3} + u | - \log | \sqrt{3} - u | \right). \end{align*}$

Putting these integrals back into our original expression we have

$\begin{align*} \int \frac{\sqrt{3-x^2}}{x} \, dx &= -\int \frac{x}{\sqrt{3-x^2}} \, dx + 3 \int \frac{1}{x \sqrt{3-x^2}} \, dx \\[10pt] &= \sqrt{3-x^2} - \frac{\sqrt{3}}{2} \left( \log | \sqrt{3} + u| - \log | \sqrt{3} - u| \right) \\[10pt] &= \sqrt{3-x^2} - \frac{\sqrt{3}}{2} \log \left| \frac{\sqrt{3} + \sqrt{3-x^2}}{\sqrt{3} - \sqrt{3-x^2}} \right| + C \\[10pt] &= \sqrt{3-x^2} - \frac{\sqrt{3}}{2} \log \left| \frac{(\sqrt{3} + \sqrt{3-x^2})^2}{x^2} \right| + C \\[10pt] &= \sqrt{3-x^2} - \sqrt{3} \log \left( \frac{\sqrt{3} + \sqrt{3-x^2}}{x} \right) + C. \end{align*}$

Evaluate the integral of 1 / (1+ a cos x) for a > 1

by RoRi

December 17, 2015

Compute the following integral.

$\int \frac{dx}{1+a \cos x} \qquad \text{for} \quad a > 1.$

Using the previous exercise we know that with the substitution $u = \tan \frac{x}{2}$ we obtain

$\int \frac{dx}{1 + a \cos x} = \frac{2}{1+a} \int \frac{du}{\left( \frac{1-a}{1+a} \right) u^2 +1}.$

This time, since $a > 1$ we make the substitution

$t = u\sqrt{\frac{a-1}{a+1}} \qquad dt = \sqrt{\frac{a-1}{a+1}} \, du.$

This gives us, for $a > 1$ ,

$\begin{align*} \int \frac{dx}{1+a \cos x} &= \frac{2}{1+a} \int \frac{du}{ \left( \frac{1-a}{1+a} \right) u^2 + 1} \\[9pt] &= \frac{-2}{1+a} \sqrt{\frac{a+1}{a-1}} \int \frac{dt}{t^2-1}. \end{align*}$

Using partial fractions we can rewrite the integrand in this to obtain,

$\begin{align*} \frac{-2}{1+a} \sqrt{\frac{a+1}{a-1}} \int \frac{dt}{t^2-1} &= \frac{-2}{\sqrt{a^2-1}} \int \left( \frac{1/2}{t-1} - \frac{1/2}{t+1} \right) \, dt \\[9pt] &= \frac{-1}{\sqrt{a^2-1}} \left( \log |t-1| - \log |t+1| \right) + C \\[9pt] &= \frac{1}{\sqrt{a^2-1}} \log \left| \frac{s+1}{s-1} \right| + C \\[9pt] &= \frac{1}{\sqrt{a^2-1}} \log \left| \frac{\sqrt{\frac{a-1}{a+1}}u + 1}{\sqrt{\frac{a-1}{a+1}} u - 1} \right| + C \\[9pt] &= \frac{1}{\sqrt{a^2-1}} \log \left| \frac{\sqrt{a-1} u + \sqrt{a+1}}{\sqrt{a-1} u - \sqrt{a+1}} \right| + C\\[9pt] &= \frac{1}{\sqrt{a^2-1}} \log \left| \frac{ \left( \sqrt{a-1} u + \sqrt{a+1} \right)^2}{(a-1)u^2 - (a+1)} \right| + C \\[9pt] &= \frac{1}{\sqrt{a^2-1}} \log \left| \frac{au^2-u^2 + 2u \sqrt{a^2-1} + a + 1}{au^2 - u^2 - a - 1} \right| + C\\[9pt] &= \frac{1}{\sqrt{a^2-1}} \log \left| \frac{a (u^2+1) - (u^2-1) + 2u\sqrt{a^2-1}}{a(u^2-1) - (u^2+1)} \right| + C \\[9pt] &= \frac{-1}{\sqrt{a^2-1}} \log \left| \frac{a + \frac{1-u^2}{1+u^2} + \frac{2u}{1+u^2} \sqrt{a^2-1}}{a \frac{1-u^2}{1+u^2} + \frac{1+u^2}{1+u^2}} \right| + C\\[9pt] &= \frac{-1}{\sqrt{a^2-1}} \log \left| \frac{a + \cos x + \sqrt{a^2-1} \sin x}{1+ a \cos x} \right| + C. \end{align*}$

Evaluate the integral of 1 / (x⁴ + 1)

by RoRi

December 4, 2015

Compute the following integral.

$\frac{dx}{x^4+1}.$

First, we need to rewrite the denominator as a product of linear and quadratic terms so that we can use partial fraction decomposition. To that end, we have

$x^4 + 1 = (x^2 - \sqrt{2} \, x + 1)(x^2 + \sqrt{2} \, x + 1).$

Then we write,

$\frac{1}{x^4+1} = \frac{Ax+B}{x^2-\sqrt{2} \, x +1} + \frac{Cx+D}{x^2 + \sqrt{2}\, x + 1}.$

This gives us the equation

$(Ax+B)(x^2+\sqrt{2} \, x + 1) + (Cx+D)(x^2 - \sqrt{2} \, x + 1) = 1.$

Here we multiply everything out and equate like powers of $x$ . Multiplying out, and grouping like powers of $x$ give us,

$(A+C) x^3 + (A\sqrt{2}+B-C\sqrt{2}+D) x^2 + (A+B\sqrt{2}+C-D\sqrt{2}) x + (B+D) = 1.$

From this we have the system of equations

$\begin{align*} A+C &= 0 \\ A\sqrt{2} + B - C \sqrt{2} + D &= 0 \\ A+B \sqrt{2} + C - D \sqrt{2} &= 0 \\ B+D &= 1. \end{align*}$

Solving this system (sorry, I’m omitting the details of solving the system… there’s nothing tricky in it, but it’s tedious to TeX up into the blog) we obtain

$A = -\frac{1}{2\sqrt{2}}, \quad B = \frac{1}{2}, \quad C = \frac{1}{2\sqrt{2}}, \quad D = \frac{1}{2}.$

Therefore, we have

$\begin{align*} \int \frac{dx}{x^4+1} &= \int \left( \frac{Ax+B}{x^2-\sqrt{2} \, x +1} + \frac{Cx+D}{x^2+\sqrt{2}\,x+1} \right) \, dx \\[10pt] &= \frac{1}{4\sqrt{2}} \left(- \int \frac{2x-\sqrt{2} \, dx}{x^2-\sqrt{2} \, x +1} + \int \frac{\sqrt{2} \, dx}{x^2-\sqrt{2}\, x+1} \right. \\ &\qquad \qquad \left.+ \int \frac{2x+\sqrt{2} \, dx}{x^2+\sqrt{2} \, x + 1} + \int \frac{\sqrt{2} \, dx}{x^2+\sqrt{2} \, x + 1}\right)\\[10pt] &= \frac{1}{4\sqrt{2}} \left( \log \left| x^2 + \sqrt{2} \, x + 1\right| - \log \left| x^2 - \sqrt{2} \, x + 1 \right| \right) \\ & \qquad \qquad + \frac{1}{4} \int \frac{dx}{x^2+\sqrt{2} \, x + 1} + \frac{1}{4} \int \frac{dx}{x^2 -\sqrt{2} \, x + 1} \\[10pt] &= \frac{1}{4\sqrt{2}} \log \left| \frac{x^2+\sqrt{2} \, x+1}{x^2-\sqrt{2} \, x +1} \right| + \frac{1}{4} \int \frac{dx}{\left(x+\frac{\sqrt{2}}{2}\right)^2 + \frac{1}{2}} + \frac{1}{4} \int \frac{dx}{\left(x-\frac{\sqrt{2}}{2}\right)^2+\frac{1}{2}} \\[10pt] &= \frac{1}{4\sqrt{2}} \log \left| \frac{x^2+\sqrt{2} \, x+1}{x^2-\sqrt{2} \, x +1} \right| + \frac{1}{2} \int \frac{dx}{(\sqrt{2}\, x+1)^2 +1} + \frac{1}{2} \int \frac{dx}{(\sqrt{2}\, x -1)^2 +1} \\[10pt] &= \frac{1}{4\sqrt{2}} \log \left| \frac{x^2+\sqrt{2} \, x+1}{x^2-\sqrt{2} \, x +1} \right| + \frac{1}{2\sqrt{2}} \int \frac{\sqrt{2} \, dx}{(\sqrt{2} \, x + 1)^2 + 1} + \frac{1}{2 \sqrt{2}} \int \frac{\sqrt{2} \, dx}{(\sqrt{2} \, x - 1)^2 + 1} \\[10pt] &= \frac{1}{4\sqrt{2}} \log \left| \frac{x^2+\sqrt{2} \, x+1}{x^2-\sqrt{2} \, x +1} \right| + \frac{1}{2 \sqrt{2}} \left( \arctan (\sqrt{2} \, x + 1) + \arctan (\sqrt{2} \, x - 1) \right) + C. \end{align*}$

Now, we want to simplify the $\arctan$ expressions (to get our answer to match the answer Apostol provided in the back of the book). So, we use the identity

$\arctan (x) + \arctan (y) = \arctan \left( \frac{x + y}{1 - xy} \right).$

Therefore,

$\begin{align*} \arctan (\sqrt{2} \, x + 1) + \arctan (\sqrt{2} \, x - 1) &= \arctan \left( \frac{\sqrt{2} \, x + 1 + \sqrt{2} \, x - 1}{1 - (\sqrt{2}\ , x +1)(\sqrt{2} \, x - 1)} \right)\\ &= \arctan \left( \frac{2\sqrt{2} \, x}{2-2x^2} \right) \\ &= \arctan \left( \frac{\sqrt{2} \, x}{1-x^2} \right). \end{align*}$

Finally, putting this back into the formula we obtained above, we have

$\int \frac{dx}{x^4 + 1} = \frac{1}{4\sqrt{2}} \log \left| \frac{x^2+\sqrt{2} \, x+1}{x^2-\sqrt{2} \, x +1} \right| + \frac{1}{2\sqrt{2}} \arctan \left( \frac{\sqrt{2} \, x}{1-x^2} \right) + C.$

Evaluate the integral of 1 / (x⁴ – 1)

by RoRi

December 4, 2015

Compute the following integral.

$\int \frac{dx}{x^4-1}.$

First, we factor the denominator

$x^4 - 1= (x^2 - 1)(x^2+1) = (x-1)(x+1)(x^2+1).$

To apply partial fraction decomposition we then write,

$\frac{1}{x^4-1} = \frac{A}{x-1} + \frac{B}{x+1} + \frac{Cx+D}{x^2+1}.$

This gives us the equation

$A(x+1)(x^2+1) + B(x-1)(x^2+1) + (Cx+D)(x-1)(x+1) = 1.$

First, we evaluate the equation at $x = 1$ and $x = -1$ to obtain

$\begin{align*} 4A &= 1 & \implies \qquad A &= \frac{1}{4} \\ -4B &= 1 & \implies \qquad B &= -\frac{1}{4}. \end{align*}$

Then, using these values of $A$ and $B$ we evaluate at $x = 0$ to obtain

$\frac{1}{4} + \frac{1}{4} - D = 1 \qquad \implies \qquad D = -\frac{1}{2}.$

Finally, equating like powers of $x$ in the equation we must have $C = 0$ . Therefore,

$\begin{align*} \int \frac{1}{x^4-1} \, dx &= \int \left( \frac{A}{x-1} + \frac{B}{x+1} + \frac{Cx+D}{x^2+1} \right) \, dx \\[9pt] &= \frac{1}{4} \int \frac{1}{x-1} \, dx - \frac{1}{4} \int \frac{1}{x+1} \, dx - \frac{1}{2} \int \frac{1}{x^2+1} \, dx \\[9pt] &= \frac{1}{4} \log |x-1| - \frac{1}{4} \log |x+1| - \frac{1}{2} \arctan x + C\\[9pt] &= \frac{1}{4} \log \left| \frac{x-1}{x+1} \right| - \frac{1}{2} \arctan x + C. \end{align*}$

Evaluate the integral of (1 – x³) / (x(x² + 1))

by RoRi

December 4, 2015

Compute the following integral.

$\int \frac{1-x^3}{x(x^2+1)} \, dx.$

First, we factor the numerator and do some rearranging of the integral,

$\begin{align*} \int \frac{1-x^3}{x(x^2+1)} \, dx &= \int \left( \frac{1}{x(x^2+1)} - \frac{x^2}{1+x^2} \right) \, dx \\[9pt] &= \int \frac{1}{x(x^2+1)} \, dx - \int \frac{x^2+1}{x^2+1} \, dx + \int \frac{1}{1+x^2} \, dx \\[9pt] &= \int \frac{1}{x(x^2+1)} \, dx - x + \arctan x + C. \end{align*}$

For the remaining integral $\int \frac{1}{x(x^2+1)} \, dx$ we use partial fractions, writing

$\frac{1}{x(x^2+1)} = \frac{A}{x} + \frac{Bx+C}{x^2+1}.$

This gives us the equation

$A(x^2+1) + (Bx+C)x = 1.$

First, we evaluate at $x = 0$ to obtain

$A = 1.$

Then we have

$x^2+1 + Bx^2 + Cx = 1 \quad \implies \quad (1+B)x^2 + Cx = 0.$

Equating like powers of $x$ we have $B = -1$ and $C = 0$ . Therefore,

$\begin{align*} \int \frac{1-x^3}{x(x^2+1)} \, dx &= \int \frac{1}{x} \, dx - \int \frac{x}{x^2+1} \ , dx - x + \arctan x + C \\[9pt] &= \log |x| - \frac{1}{2} \log (x^2+1) - x + \arctan x + C \\[9pt] &= \log \frac{|x|}{\sqrt{x^2+1}} - x + \arctan x + C. \end{align*}$

Evaluate the integral of 1 / (x⁴ – 2x³)

by RoRi

December 4, 2015

Compute the following integral.

$\int \frac{dx}{x^4 - 2x^3}.$

First, we want to use partial fraction decomposition so we write

$\frac{1}{x^4-2x^3} = \frac{1}{x^3 (x - 2)} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x^3} + \frac{D}{x-2}.$

This gives us the equation

$Ax^2(x-2) + Bx(x-2) + C(x-2) + Dx^3 = 1.$

Evaluating at $x = 0$ and $x = 2$ we obtain the values of $C$ and $D$ ,

$\begin{align*} -2C &= 1 &\implies \qquad C &= -\frac{1}{2} \\ 8D &= 1 & \implies \qquad D &= \frac{1}{8}. \end{align*}$

Using these values of $C$ and $D$ and evaluating at $x = 1$ and $x = -1$ we obtain

$\begin{align*} -A - B + \frac{1}{2} + \frac{1}{8} &= 1 \\ -3A + 3B + \frac{3}{2} - \frac{1}{8} &= 1. \end{align*}$

Solving this system for $A$ and $B$ we obtain

$A = -\frac{1}{8}, \qquad B = -\frac{1}{4}.$

Therefore,

$\begin{align*} \int \frac{1}{x^4-2x^3} \, dx &= \int \left( \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x^3} + \frac{D}{x-2} \right) \, dx \\[9pt] &= -\frac{1}{8} \int \frac{1}{x} \, dx - \frac{1}{4} \int \frac{1}{x^2} \, dx -\frac{1}{2} \int \frac{1}{x^3} \,dx + \frac{1}{8} \int \frac{1}{x-2} \, dx \\[9pt] &= -\frac{1}{8} \log |x| + \frac{1}{4x} + \frac{1}{4x^2} + \frac{1}{8} \log |x-2| + C \\[9pt] &= \frac{1}{4x} + \frac{1}{4x^2} + \frac{1}{8} \log \left| \frac{x-2}{x} \right| + C. \end{align*}$

Evaluate the integral of (x⁴ + 1) / (x (x² + 1)²)

by RoRi

December 4, 2015

Compute the following integral.

$\int \frac{x^4 + 1}{x(x^2+1)^2} \, dx.$

The denominator is already factored, so we write

$\frac{x^4+1}{x(x^2+1)^2} = \frac{A}{x} + \frac{Bx+C}{x^2+1} + \frac{Dx+E}{(x^2+1)^2}.$

This gives us the equation

$A(x^2+1)^2 + (Bx+C)(x)(x^2+1) + (Dx+E)x = x^4 + 1.$

First, we can evaluate at $x = 0$ to obtain

$A = 1.$

Then, we multiply out and equate like powers of $x$ ,

$\begin{align*} &&(x^2+1)^2 + (Bx+C)(x)(x^2+1) + (Dx+E)x &= x^4 + 1 \\ \implies && x^4 + 2x^2 + 1 + Bx^4 + Cx^3 + Bx^2 + Cx + Dx^2 + Ex &= x^4 + 1 \\ \implies && (1+B)x^4 + Cx^3 + (2 + B + D)x^2 + (C + E)x &= x^4. \end{align*}$

Therefore, equating like powers of $x$ , we have the following equations

$\begin{align*} 1+B &= 1 \\ C &= 0 \\ 2+B+D &= 0 \\ C+E &= 0 \end{align*}$

Solving this system we obtain

$B = 0, \quad C = 0, \quad D = -2, \quad E = 0.$

Therefore, evaluating the integral we have

$\begin{align*} \int \frac{x^4+1}{x(x^2+1)^2} \, dx &= \int \left( \frac{A}{x} + \frac{Bx+C}{x^2+1} + \frac{Dx+E}{(x^2+1)^2} \right) \, dx \\[9pt] &= \int \frac{1}{x} \, dx - \int \frac{2x}{(x^2+1)^2} \, dx \\[9pt] &= \log |x| + \frac{1}{x^2+1} + C. \end{align*}$

Evaluate the integral of (x+1) / (x³-1)

by RoRi

December 4, 2015

Compute the following integral.

$\int \frac{x+1}{x^3-1} \, dx.$

Since the denominator factors as

$x^3 - 1 = (x-1)(x^2 + x + 1)$

we use partial fractions and write

$\frac{x+1}{x^3-1} = \frac{x+1}{(x-1)(x^2+x+1)} = \frac{A}{x-1} + \frac{Bx+C}{x^2+x+1}.$

This gives us the equation

$A(x^2+x+1) + (Bx+C)(x-1) = x+1.$

First, we evaluate at $x = 1$ to find the value of $A$ ,

$3A = 2 \quad \implies \quad A = \frac{2}{3}.$

Then, using this value of $A$ and evaluating at $x = 0$ to obtain the value of $C$ ,

$\frac{2}{3} - C = 1 \qquad \implies \qquad C = -\frac{1}{3}.$

Finally, we evaluate at $x = 2$ and use the values of $A$ and $C$ to compute $B$ ,

$\frac{14}{3} + 2B -\frac{1}{3} = 3 \qquad \implies \qquad B = -\frac{2}{3}.$

Therefore,

$\begin{align*} \int \frac{x+1}{x^3-1} \, dx &= \int \frac{x+1}{(x-1)(x^2+x+1)} \, dx \\[9pt] &= \int \left( \frac{A}{x-1} + \frac{Bx+C}{x^2+x+1} \right) \, dx \\[9pt] &= \frac{2}{3} \int \frac{1}{x-1} \, dx - \frac{1}{3} \int \frac{2x+1}{x^2+x+1} \, dx \\[9pt] &= \frac{2}{3} \log |x-1| - \frac{1}{3} \log(x^2+x+1) + C \\[9pt] &= \frac{1}{3} \log \left(\frac{(x-1)^2}{x^2+x+1}\right) + C. \end{align*}$