Tag: Partial Fractions

Evaluate the integral of 1 / (x² – 1)²

by RoRi

December 4, 2015

Compute the following integral.

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

First, we write

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

Then we use partial fractions, writing

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

This gives us the equation

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

Evaluating at $x = -1$ and $x = 1$ we obtain the values for $B$ and $D$ ,

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

Using these values of $B$ and $D$ and evaluating at $x = 0$ and $x = 2$ we have

$\begin{align*} A + \frac{1}{4} - C + \frac{1}{4} &= 1 \\ 3A + \frac{1}{4} + 9C + \frac{9}{4} &= 1. \end{align*}$

Solving this system for $A$ and $C$ we get

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

Thus,

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

Evaluate the integral of (x-3) / (x³ + 3x² + 2x)

by RoRi

December 3, 2015

Compute the following integral.

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

The denominator of the integrand factors as

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

Therefore, we can use partial fractions as follows,

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

This gives us the equation

$A(x+1)(x+2) + B(x)(x+2) + C(x)(x+1) = x-3.$

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

$\begin{align*} 2A &= -3 & \implies \qquad A &= -\frac{3}{2} \\ -B &= -4 & \implies \qquad B &= 4 \\ 2C &= -5 & \implies \qquad C &= -\frac{5}{2}. \end{align*}$

Therefore, we have

$\begin{align*} \int \frac{x-3}{x^3 + 3x^2 + 2x} \, dx &= \int \left( \frac{A}{x} + \frac{B}{x+1} + \frac{C}{x+2} \right) \, dx \\[10pt] &= -\frac{3}{2} \int \frac{1}{x} \, dx + 4 \int \frac{1}{x+1} \, dx - \frac{5}{2} \int \frac{1}{x+2} \, dx \\[10pt] &= -\frac{3}{2} \log |x| + 4 \log |x+1| - \frac{5}{2} \log |x+2| + C \\[10pt] &= 4 \log |x+1| - \frac{3}{2} \log |x| - \frac{5}{2} \log |x+2| + C. \end{align*}$

Evaluate the integral of 1 / ((x² – 4x + 4)(x² – 4x + 5))

by RoRi

December 3, 2015

Compute the following integral.

$\int \frac{dx}{(x^2 - 4x + 4)(x^2 - 4x + 5)}.$

In the denominator we have

$(x^2 - 4x + 4)(x^2 - 4x + 5) = (x-2)^2 (x^2 - 4x + 5).$

Then we use partial fractions,

$\frac{1}{(x-2)^2 (x^2 - 4x + 5)} = \frac{A}{x-2} + \frac{B}{(x-2)^2} + \frac{Cx + D}{x^2 - 4x + 5}.$

This gives us the equation

$A(x-2)(x^2 - 4x + 5) + B(x^2 - 4x + 5) + (Cx + D)(x-2)^2 = 1.$

We evaluate at $x = 2$ to obtain a value for $B$ ,

$B = 1.$

Then using this value of $B$ and evaluating at $x = -1$ , $x = 0$ and $x = 1$ to obtain

$\begin{align*} -30A + 10 - 9C + 9D &= 1 \\ -10A + 5 + 4D &= 1 \\ -2A + 2 + C + D &= 1. \end{align*}$

Solving this system of equations we obtain

$A = 0, \qquad C = 0, \qquad D = -1.$

Therefore, we have

$\begin{align*} \int \frac{dx}{(x^2 - 4x + 4)(x^2 - 4x + 5)} &= \int \left( \frac{A}{x-2} + \frac{B}{(x-2)^2} + \frac{Cx+D}{x^2 - 4x + 5} \right) \, dx \\[10pt] &= \int \frac{1}{(x-2)^2} \, dx - \int \frac{1}{x^2 - 4x + 5} \, dx \\[10pt] &= \frac{-1}{x-2} - \int \frac{1}{(x-2)^2 + 1} \, dx \\[10pt] &= \frac{-1}{x-2} - \arctan (x-2) + C. \end{align*}$

Evaluate the integral of x² / (x² + x – 6)

by RoRi

December 3, 2015

Compute the following integral.

$\int \frac{x^2}{x^2 + x - 6} \, dx.$

First, we have

$\frac{x^2}{x^2 + x - 6} = 1 - \frac{x-6}{x^2 + x - 6} = 1 - \frac{x-6}{(x+3)(x-2)}.$

Therefore,

$\begin{align*} \int \frac{x^2}{x^2 + x - 6} &= \int \left( 1 - \frac{x-6}{(x+3)(x-2)} \right) \, dx \\ &= x - \int \frac{x-6}{(x+3)(x-2)} \, dx. \end{align*}$

We use partial fractions to evaluate the integral on the right. To that end, we write

$\frac{x-6}{(x+3)(x-2)} = \frac{A}{x+3} + \frac{B}{x-2}.$

This gives us the equation

$A(x-2) + B(x+3) = x-6.$

Evaluating at $x = -3$ and $x = 2$ we then have

$\begin{align*} -5A &= -9 & \implies \qquad A &= \frac{9}{5} \\ 5B &= -4 & \implies \qquad B &= -\frac{4}{5}. \end{align*}$

Therefore,

$\begin{align*} \int \frac{x^2}{x^2 + x - 6} \, dx &= x - \int \frac{x-6}{(x+3)(x-2)} \, dx \\ &= x - \frac{9}{5} \int \frac{1}{x+3} \, dx + \frac{4}{5} \int \frac{1}{x-2} \, dx \\ &= x - \frac{9}{5} \log |x+3| + \frac{4}{5} \log |x-2| + C \\ &= x + \frac{4}{5} \log |x-2| - \frac{9}{5} \log |x+3| + C. \end{align*}$

Evaluate the integral of 1 / (x³ – x)

by RoRi

December 3, 2015

Compute the following integral.

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

Since $x^3 - x$ factors as $x(x^2-1) = x(x+1)(x-1)$ we have the following,

$\frac{1}{x^3 - x} = \frac{A}{x} + \frac{B}{x+1} + \frac{C}{x-1}.$

This gives us the equation

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

Substituting the values $x = 0$ , $x = -1$ and $x = 1$ we obtain

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

Therefore, we have

$\begin{align*} \int \frac{dx}{x^3-x} &= \int \frac{-1}{x} \, dx + \frac{1}{2} \int \frac{1}{x+1} \, dx + \frac{1}{2} \int \frac{1}{x-1} \, dx \\ &= -\log|x| + \frac{1}{2} \log|x+1| + \frac{1}{2} \log|x-1| + C \\ &= \frac{1}{2} \log |x^2-1| - \log |x| + C. \end{align*}$

Evaluate the following integral 1 / ((x+1)(x+2)²(x+3)³)

by RoRi

December 3, 2015

Compute the following integral.

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

First, we use partial fraction decomposition. We write

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

This gives us the equation

$\begin{align*} A&(x+2)^2 (x+3)^3 + B(x+1)(x+2)(x+3)^3 + C(x+1)(x+3)^3 \\&+ D(x+1)(x+2)^2(x+3)^2 + E(x+1)(x+2)^2(x+3) + F(x+1)(x+2)^2 = 1. \end{align*}$

First, we substitute the values $x = -1$ , $x = -2$ , and $x = -3$ which gives us

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

Substituting these values of $A, C$ , and $F$ into our equation we have

$\begin{align*} \frac{1}{8}&(x+2)^2(x+3)^3 + B(x+1)(x+2)(x+3)^3 - (x+1)(x+3)^3 \\ &+ D(x+1)(x+2)^2(x+3)^2 + E(x+1)(x+2)^2(x+3) - \frac{1}{2}(x+1)(x+2)^2 = 1. \end{align*}$

Now, we substitute the values $x = 0$ , $x = 1$ and $x = 2$ to obtain the equations

$\begin{align*} \frac{27}{4} + 54B - 27 + 36D + 12E - 2 &= 1 \\ 72 + 384B - 128 + 288D + 72E - 9 &= 1 \\ 250 + 1500B - 375 + 1200D + 240E - 24 &= 1. \end{align*}$

Solving this system we obtain the values

$B = 2, \qquad D = -\frac{17}{8}, \qquad E = -\frac{5}{4}.$

Therefore, we have the following,

$\begin{align*} \int \frac{dx}{(x+1)(x+2)^2(x+3)^3} &= \int \left( \frac{A}{x+1} + \frac{B}{x+2} + \frac{C}{(x+2)^2} \right.\\ & \left. \qquad + \frac{D}{x+3} + \frac{E}{(x+3)^2} + \frac{F}{(x+3)^3} \right) \, dx \\[10pt] &= \frac{1}{8} \int \frac{dx}{x+1} + 2 \int \frac{dx}{x+2} - \int \frac{dx}{(x+2)^2} \\ & \qquad -\frac{17}{8} \int \frac{dx}{x+3} - \frac{5}{4} \int \frac{dx}{(x+3)^2} - \frac{1}{2} \int \frac{dx}{(x+3)^3} \\[10pt] &= \frac{1}{8}\log |x+1| + 2 \log |x+2| + \frac{1}{x+2} - \frac{17}{8} \log |x+3| \\ & \qquad + \frac{5}{4} \frac{1}{x+3} + \frac{1}{4} \frac{1}{(x+3)^2} + C \\[10pt] &= \frac{4(x+3)^2 + 5(x+2)(x+3) + x + 2 }{4 (x+2)(x+3)^2} \\ & \qquad + \frac{1}{8} \Big( \log|x+1| + 16 \log|x+2| - 17 \log|x+3| \Big) + C\\[10pt] &= \frac{9x^2 + 50x + 68}{4(x+2)(x+3)^2} + \frac{1}{8} \log \left| \frac{(x+1)(x+2)^{16}}{(x+3)^{17}} \right| + C. \end{align*}$

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

by RoRi

December 3, 2015

Compute the following integral.

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

To evaluate this we use partial fractions. First, we write

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

This gives us the equation

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

First, we evaluate at $x = 0$ to find that $A = 1$ . Using this value of $A$ we obtain the equation

$x^4 + 2x^2 + 1 + Bx^4 + Bx^2 + Cx^3 + Cx + Dx^2 +Ex = 1.$

Equating like powers of $x$ this gives us the system of equations:

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

We solve this system to obtain,

$B = -1, \quad C = 0, \quad D = -1, \quad E = 0.$

Therefore,

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

Evaluate the following integral (x+2) / (x² + x)

by RoRi

December 3, 2015

Compute the following integral.

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

We have

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

To evaluate the second integral on the right we use partial fractions. Write,

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

This gives us the equations

$A(x+1) + Bx = 1.$

Letting $x = 0$ and $x = -1$ we have

$A = 1 \qquad \text{and} \qquad B = -1.$

Therefore, we have

$\begin{align*} \int \frac{x+2}{x^2+x} \, dx &= \int \frac{1}{x} \, dx + \int \frac{1}{x(x+1)} \, dx \\ &= \log |x| + \int \frac{1}{x} \, dx - \int \frac{1}{x+1} \, dx \\ &= \log |x| + \log |x| - \log |x+1| + C \\ &= 2 \log |x| - \log|x+1| + C. \end{align*}$

Evaluate the integral of x⁴ / (x⁴ +5x² + 4)

by RoRi

December 3, 2015

Compute the following integral.

$\int \frac{x^4 \, dx}{x^4 + 5x^2 + 4}.$

First, we simplify the integrand and then use partial fractions,

$\int \frac{x^4 \, dx}{x^4 + 5x^2 + 4} = \int \frac{x^4 + 5x^2 + 4 - 5x^2 - 4}{x^4 + 5x^2 + 4} \, dx = \int \, dx - \int \frac{5x^2 + 4}{x^4 + 5x^2 + 4} \, dx.$

To evaluate the second integral we use partial fractions. We have

$x^4 + 5x^2 + 4 = (x^2+4)(x^2+1).$

Therefore, we write

$\frac{5x^2+4}{x^4 + 5x^2 + 4} = \frac{Ax+B}{x^2+4} + \frac{Cx + D}{x^2+1}.$

Then we have the equation

$\begin{align*} &&(Ax+B)(x^2+1) + (Cx+D)(x^2+4) &= 5x^2 + 4 \\ \implies && (A+C)x^3 + (B+D)x^2 + (A+4C)x + (B+4D) &= 5x^2 + 4. \end{align*}$

Equating like powers of $x$ we then have four equations and four unknowns:

$\begin{align*} A+C &= 0 \\ B+D &= 5 \\ A+4C &= 0 \\ B+4D &= 4. \end{align*}$

Solving this system we find $A = C = 0$ , $B = \frac{16}{3}$ and $D = -\frac{1}{3}$ . Therefore, we have

$\begin{align*} \int \frac{x^4 \, dx}{x^4 + 5x^2 + 4} &= \int dx - \int \frac{5x^2+4}{(x^2+4)(x^2+1)} \, dx \\ &= x - \frac{16}{3} \int \frac{dx}{x^2+4} + \frac{1}{3} \int \frac{dx}{1+x^2} \\ &= x + \frac{1}{3} \arctan x - \frac{8}{3} \arctan \left( \frac{x}{2} \right) + C. \end{align*}$

Evaluate the integral of (8x³ + 7) / ((x+1)(2x+1)³)

by RoRi

December 3, 2015

Compute the following integral.

$\int \frac{8x^3 + 7}{(x+1)(2x+1)^3} \, dx.$

Since the denominator is already factored into linear terms we can proceed directly with the partial fraction decomposition. We write,

$\frac{8x^3 + 7}{(x+1)(2x+1)^3} = \frac{A}{x+1} + \frac{B}{2x+1} + \frac{C}{(2x+1)^2} + \frac{D}{(2x+1)^3}.$

This gives us the equation

$A(2x+1)^3 + B(x+1)(2x+1)^2 + C(x+1)(2x+1) + D(x+1) = 8x^3 + 7.$

First, we can find the values of $A$ and $D$ by evaluating at $x = -1$ and $x = -\frac{1}{2}$ , respectively. This gives us

$\begin{align*} -A &= -1 & \implies \quad A &= 1 \\ \frac{1}{2}D &= 6 & \implies \quad D &= 12. \end{align*}$

Then using these values of $A$ and $D$ we evaluate at $x = 0$ and $x = 1$ (these are just convenient values, there isn’t anything special about them) to obtain the two equations

$\begin{align*} 1 + B + C + 12 &= 7 & \implies \quad B + C &= -6 \\ 27 + 18B + 6C + 24 &= 15 & \implies \quad 3B + C &= -6. \end{align*}$

Solving these two equations we obtain $B = 0$ and $C = -6$ . Therefore, we have the following partial fraction decomposition:

$\frac{8x^3 + 7}{(x+1)(2x+1)^3} = \frac{1}{x+1} - \frac{6}{(2x+1)^2} + \frac{12}{(2x+1)^3}.$

We can now evaluate the integral,

$\begin{align*} \int \frac{8x^3 + 7}{(x+1)(2x+1)^3} \, dx &= \int \frac{dx}{x+1} - 3 \int \frac{2 \, dx}{(2x+1)^2} + 6 \int \frac{2\, dx}{(2x+1)^3} \\ &= \log |x+1| + \frac{3}{2x+1} - \frac{3}{(2x+1)^2} + C. \end{align*}$