Integrating Rational Functions with Substitution

[binbagsss]

Homework Statement

Apologies if this is obvious, maybe I'm a little out of touch

$\int\limits^b_0 \frac{x^3}{x^2+m^2} dx$

Homework Equations

The Attempt at a Solution

I [/B]was going to go by parts breaking the $x^3 = x^2 . x$

So that I have the logarithm

I.e :

$b^2 \frac{log (b^2 + m^2)}{2} - \int \frac{log (x^2+m^2)}{2} 2x dx$
But the solution is :

$b^2 + m^2 log ( \frac{m^2}{b^2+m^2} )$( I thought that perhaps the solution could be going by parts again, but there is no reason for the boundary term to vanish )

Ta

 

[Mark44]

Homework Statement

Apologies if this is obvious, maybe I'm a little out of touch

$\int\limits^b_0 \frac{x^3}{x^2+m^2} dx$

Homework Equations

The Attempt at a Solution

I [/B]was going to go by parts breaking the $x^3 = x^2 . x$

So that I have the logarithm

I.e :

$b^2 \frac{log (b^2 + m^2)}{2} - \int \frac{log (x^2+m^2)}{2} 2x dx$
But the solution is :

$b^2 + m^2 log ( \frac{m^2}{b^2+m^2} )$( I thought that perhaps the solution could be going by parts again, but there is no reason for the boundary term to vanish )

Ta

The second term in your answer can be integrated using substitution, with $u = x^2 + b^2, du = 2xdx$.

 

[Delta2]

Write $x^3=x(x^2+m^2)-m^2x$ so the integral becomes $\int xdx-\int \frac{m^2x}{x^2+m^2}dx$.

 

[vela]

Or just use the substitution $u=x^2+m^2$ on the original integral.