Question on integration technique

[stickyhub91]

[attachment=0:3qm3i9fs]calc.jpg[/attachment:3qm3i9fs]

How do I integrate this question?

 

[Ryan Rigdon]

hope this helps. oops i saw a mistake be right back. made the corrections.

 

[stickyhub91]

Ahh.. got it now!

 

[Ryan Rigdon]

glad to help.

 

[BigGlenntheHeavy]

\(\displaystyle \int\frac{dx}{(x+1)^2(x^2+2x)^{1/2}}, \ Let \ u \ = \ x+1. \ du \ = \ dx, \ and \ x \ = \ u-1.\)

\(\displaystyle So, \ \int\frac{du}{u^2[(u-1)^2+2(u-1)]^{1/2}} \ = \ \int\frac{du}{u^2(u^2-1)^{1/2}}\)

\(\displaystyle Now \ let \ u \ = \ sec(a), \ du \ = \ sec(a)tan(a)da.\)

\(\displaystyle \int\frac{sec(a)tan(a)da}{sec^2(a)[sec^2(a)-1]^{1/2}} \ = \ \int\frac{sec(a)tan(a)da}{sec^2(a)tan(a)}\)

\(\displaystyle = \ \int\frac{da}{sec(a)} \ = \ \int cos(a)da \ = \ sin(a)+C\)

\(\displaystyle Now, \ sec(a) \ = \ u, \ hence \ sin(a) \ = \ \frac{\sqrt{u^2-1}}{u} \ = \ (u \ = \ x+1) \ = \ \frac{\sqrt{(x+1)^2-1}}{x+1}\)

\(\displaystyle Therefore \ \int\frac{dx}{(x+1)^2(x^2+2x)^{1/2}} \ = \ \frac{\sqrt{x^2+2x}}{x+1}+C\)