trig substitution again: integral of 1/sqrt(x^2+4x) dx

[cheffy]

integral of 1/sqrt(x^2+4x) dx

I did a trig substitution of x=2tan(y), but I got to integral of 1/sqrt(u^2+u) du where u=tan(y), but now I'm stuck/I'm not sure if what I did was right. Any suggestions? Thanks!

 

[galactus]

One way is to let \(\displaystyle x=u-2, \;\ dx=du\)

That gives:

\(\displaystyle \L\\\int\frac{1}{\sqrt{(u-2)^{2}+4(u-2)}}du\)

Which simplifies to:

\(\displaystyle \L\\\int\frac{1}{\sqrt{u^{2}-4}}du\)

Now, it's a little easier to integrate.