Integral of x/(x^2+z^2)^(3/2) from 0 to a

[klawlor419]

$$\int_{0}^{a}\frac{x}{(x^2+z^2)^\frac{3}{2}}dx$$

Hi all, I'm stuck on this one. Sure there's an easier way to do it. Don't have my calc book currently. Thanks!

 

[Adithyan]

Put x² = t.

 

[klawlor419]

Got it with the trig sub $$x=z \tan{\theta}$$
Thanks!

 

[Adithyan]

Got it with the trig sub $$x=z \tan{\theta}$$
Thanks!

No, that substitution isn't convenient,

If you substitute x² = t, you would get the numerator as dt (since dt= 2xdx)

 

[klawlor419]

Thanks! That is slightly easier!

 

[HallsofIvy]

I would have used $t= x^2+ z^2$ so that $dt= 2x dx$, $(1/2)dt= xdx$.

The integral becomes
$$\frac{1}{2}\int_{z^2}^{z^2+ a^2} t^{-3/2}dt$$

 

[klawlor419]

Thanks! Thats slightly easier still!