Integral of sqrt((7-x)/(6+x))

[garfieldtom]

Hi, Can someone kindly tell me the approach I should take to solve the integral of sqrt((7-x)/(6+x))? At first, looking at the square root I thought it's trig substitution, but then there's no x^2 anywhere inside the square root. So I don't know how to start....Any pointers will be appreciated!!

 

[mmm4444bot]

Hello Garfield:

My integral calculus is rusty.

If you multiply the numerator and denominator by the conjugate of the denominator, then you'll have 36 - x^2 on the bottom.

Does that help?

Gobble, gobble,

~ Mark

 

[galactus]

Just another suggestion. There are many ways to look at something like this.

But if we let \(\displaystyle u=\sqrt{6-x}\), we get it whittled down to:

\(\displaystyle -2\int\sqrt{13-u^{2}}du\)

Now, trig sub is handy to use.