How to evaluate:#intx^5/(x^4 + 1)^2dx#?

Question

Charles Anctil · Accepted Answer

#1/4tan^-1(x^2)-1/4 x^2/(x^4+1)+C# Substitute #x^2=tan theta# in this, with #2xdx = sec^2theta d theta# to get # int x^5/(x^4+1)^2 dx = 1/2 int x^4/(x^4+1)^2 2x dx# #qquad = 1/2 int tan^2 theta/(tan^2 theta+1)^2 sec^2 theta d theta = 1/2 int (tan^2 theta sec^2 theta)/sec^4 theta d theta# #qquad = 1/2int sin^2 theta d theta = 1/4 int (1-cos(2 theta)) d theta # #qquad = 1/4 theta -1/8 sin(2 theta) = theta/4 -1/4 sin theta cos theta# #qquad = theta/4 -1/4 tan theta/sec^2 theta = 1/4tan^-1(x^2)-1/4 x^2/(x^4+1)+C#

Luke Alvarez · Answer

undefined To evaluate the integral $ \int \frac{x^5}{(x^4 + 1)^2} \, dx $, we can use a substitution method. Let $ u = x^4 + 1 $. Then, $ du = 4x^3 \, dx $. Rearranging for $ dx $, we get $ dx = \frac{du}{4x^3} $.

Substitute $ u = x^4 + 1 $ and $ dx = \frac{du}{4x^3} $ into the integral:

$$ \int \frac{x^5}{(x^4 + 1)^2} \, dx = \int \frac{x^5}{u^2} \cdot \frac{du}{4x^3} = \frac{1}{4} \int \frac{1}{u^2} \, du $$

This simplifies to:

$$ \frac{1}{4} \int u^{-2} \, du = \frac{1}{4} \cdot \frac{u^{-1}}{-1} + C $$

Substitute back $ u = x^4 + 1 $:

$$ = -\frac{1}{4(x^4 + 1)} + C $$

So, the evaluated integral is $ -\frac{1}{4(x^4 + 1)} + C $.