Compute the following integral.

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

Then we write,

This gives us the equation

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

From this we have the system of equations

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

Therefore, we have

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

Therefore,

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

The “identity” of arctan(x) + arctan(y) is false, arctan(√2x+1) + arctan(√2x-1) has the same derivative as arctan(√2x/(1-x²)) over three intervals but not the entire line

Here is shorter evaluation of the integral

I = integral 1/(1+x^4) dx

= 1/2 * 2/(1+x^4) dx

= 1/2 *( x^2 + 1 – (x^2-1))/(1+x^4) dx

= 1/2 ( (x^2+1)/(x^4+1) – (x^2-1)/(1+x^4) dx

Let I1 and I2 be the two integrals on the right

I1 = int (x^2+1)/( x^4+1) dx

= ( 1+1/x^2)/( x^2 + 1/x^2 ) dx

= ( 1 + 1/x^2)/( (x-1/x)^2 + 2) dx

let u = x – 1/x

du = (1 + 1/x^2)dx

I1 = du/( u^2 + 2)

= arctan(u/sqrt(2)) / sqrt(2)

= arctan( (x-1/x)/sqrt(2))/sqrt(2)

I2 = int (x^2-1)/( x^4+1) dx

= ( 1-1/x^2)/( x^2 + 1/x^2 ) dx

= ( 1 + 1/x^2)/( (x+1/x)^2 – 2) dx

let u = x + 1/x

du = (1 – 1/x^2)dx

I2 = du/( u^2 – 2)

= arctanh(u/sqrt(2)) / sqrt(2)

= -arctanh( (x+1/x)/sqrt(2))/sqrt(2)

I = 1/2*( arctan( (x-1/x)/sqrt(2))/sqrt(2) + arctanh( (x+1/x)/sqrt(2))/sqrt(2))