Solving Difficult Integral: Help Appreciated!

[Amad27]

Hello,

$\int \frac{x}{(x^2+1)(bx+1)}$

Is the problem, which I am trying to work through. "b" is just a constant, it is very tough to integrate this indefinitely, any help will be greatly appreciated. Off the top of this,

I tried with the u-substitution,

let $u = bx + 1$ and $x = (u-1)/b$
$x^2 = [(u-1)/b]^2$
$du = b$

= $(b)\int \frac{\frac{u-1}{b}}{\frac{u((u-1)^2+b^2)}{b^2}} du$

= $(b)\int \frac{b(u-1)}{u[(u-1)^2+b^2]}$

This is where I get lost.. Any help? Thanks

 

[da_nang]

Have you tried partial fraction decomposition?

 

[chingel]

Yes try to do partial fractions, it will work out.

 

[Amad27]

Have you tried partial fraction decomposition?

Yes, look below

 

[benorin]

Partial Fraction Decomposition

Set $\frac{x}{(x^2+1)(bx+1)}=\frac{Ax+B}{x^2+1}+\frac{C}{bx+1}\Rightarrow x= (Ax+B) (bx+1)+C(x^2+1)\Rightarrow x=(Ab+C)x^2+(A+Bb)x+(B+C)$

so that we have the equations (by equating coefficients of like powers of x): $Ab+C=0,A+Bb=1,B+C=0\Rightarrow A=1/(b^2+1),B=b/(b^2+1),C=-b/(b^2+1)$

hence $\frac{x}{(x^2+1)(bx+1)}=\frac{1}{b^2+1}\left(\frac{x+b}{x^2+1}-\frac{b}{bx+1}\right)$

 

[Amad27]

Set $\frac{x}{(x^2+1)(bx+1)}=\frac{Ax+B}{x^2+1}+\frac{C}{bx+1}\Rightarrow x= (Ax+B) (bx+1)+C(x^2+1)\Rightarrow x=(Ab+C)x^2+(A+Bb)x+(B+C)$

so that we have the equations (by equating coefficients of like powers of x): $Ab+C=0,A+Bb=1,B+C=0\Rightarrow A=1/(b^2+1),B=b/(b^2+1),C=-b/(b^2+1)$

hence $\frac{x}{(x^2+1)(bx+1)}=\frac{1}{b^2+1}\left(\frac{x+b}{x^2+1}-\frac{b}{bx+1}\right)$

Hello,

Yes, I got it. I got the correct answer; this was part of a Putnam integral by the way.This thread can now be locked.