Solving Integral of (u-2)/sqrt(u^2+1) du - Help!

[sony]

Ok so after a substituion I have this: (I don't know how to proceed)

"integral of" (u-2) / sqrt(u^2 + 1) du

and by the way, can someone tell me what cot^3(tan^-1(x)) equals?

Thanks!

 

[tmc]

sqrt(u^2+1) just screams for a trig substitution.

 

[sony]

Bah, I see it. I easily get confused when having to substitution i several steps :P

Thanks

 

[TD]

Well, it may be a bit easier if you split at first.

$$\int {\frac{{u - 2}}{{\sqrt {u^2 + 1} }}du} = \int {\frac{u}{{\sqrt {u^2 + 1} }}du} - \int {\frac{2}{{\sqrt {u^2 + 1} }}du}$$

So the first one gives

$$\int {\frac{u}{{\sqrt {u^2 + 1} }}du} = \frac{1}{2}\int {\frac{{d\left( {u^2 + 1} \right)}}{{\sqrt {u^2 + 1} }}}$$

Which is easy to integrate, without a trig subst.
Then of course, still the second part

 

[hotvette]

You might want to break it up into 2 integrals. The first one is very straightforward, the second one is easy with a trig substitution as tmc suggested.

EDIT: near simultaneous post w/ TD

 

[sony]

Ok, thanks. I have that one now.

Can you please hint me with one more:
I have the integral of (sqrt(1-x^2)) / x^4 dx

Which with trig.subs leads to:
integral of cot^2(y) *csc^2(y) dy

Again I'm stuck :P

 

[TD]

Correct, but remember that

$$\left( {\cot x} \right)^\prime = - \csc ^2 x$$