Solution to ∫ dx x^2 /(1+x^2): Step-by-Step Guide

[physfed]

I would like to know the step by step solution of this integral:

∫ dx x^2 /( 1+x^2)

I tried to solve it integrating by parts with u = x dv =x /(1+x^2) , or with hyperbolic functions, but I always get stuck...

Thank you

 

[Greg Bernhardt]

I tried to solve it integrating by parts with u = x dv =x /(1+x^2) , or with hyperbolic functions, but I always get stuck...

Can you show us where you get stuck?

 

[physfed]

∫ dx x^2 /( 1+x^2)

u = x dv =x /(1+x^2)

so

∫ dx x^2 /( 1+x^2) = x log(1+x^2) -∫ dx log(1+x^2) ...stop

I also tried

u = x^2 dv = 1 /(1+x^2)

so

∫ dx x^2 /( 1+x^2) = x^2 atan(x) -∫ dx 2x atan(x) ...stop

 

[PeroK]

∫ dx x^2 /( 1+x^2)

u = x dv =x /(1+x^2)

so

∫ dx x^2 /( 1+x^2) = x log(1+x^2) -∫ dx log(1+x^2) ...stop

I also tried

u = x^2 dv = 1 /(1+x^2)

so

∫ dx x^2 /( 1+x^2) = x^2 atan(x) -∫ dx 2x atan(x) ...stop

Try $x = tan(u)$

 

[Mark44]

I would like to know the step by step solution of this integral:

∫ dx x^2 /( 1+x^2)

The integrand is an improper rational expression (degree of numerator = degree of denominator).

You can either use polynomial long division to get a proper rational expression, or do the following, which is easier:
$$\int \frac{x^2~dx}{1 + x^2} = \int \frac{1 + x^2 - 1~dx}{1 + x^2}$$
Now split into two integrals, one of which is trivial and the other you probably already know.

I tried to solve it integrating by parts with u = x dv =x /(1+x^2) , or with hyperbolic functions, but I always get stuck...

Thank you