1. ## integrating help

$\displaystyle \int\limits_0^1\frac{dx}{(1+x)\sqrt{8x^2+14x+5}}$

with substitution

2. Originally Posted by stud_02
$\displaystyle \int\limits_0^1\frac{dx}{(1+x)\sqrt{8x^2+14x+5}}$

with substitution
Try rewriting the square root in the denominator with $\displaystyle (x+1)$.

$\displaystyle 8x^2+14x+5 = 8(x+1)^2-2(x+1)-1$

Then use $\displaystyle u=x+1$ and $\displaystyle du=dx$

Now you have $\displaystyle \int_0^1\frac{1}{u\sqrt{8u^2-2u-1}}\,du$

This should make it a little easier, though to be honest I'm not sure where to go from here. I'll edit in additional steps if I think of them.

If you want it, the final answer (according to Maple) is:
Spoiler:
$\displaystyle \int\frac{1}{(1+x)\sqrt{8x^2+14x+5}}\,dx = \arctan\left(\frac{-x-2}{\sqrt{8x^2+14x+5}}\right)$

3. Sub sqrt(8x^2 + 14x + 5) + x = u , It is known as Euler's Substitution

4. starting from redsox's...

$\large \int_{0}^{1}\frac{1}{u\sqrt{8u^{2}-2u-1}}du$

$\large \int_{0}^{1}\frac{1}{2\sqrt{2}u\sqrt{u^{2}-u/4-1/8}}du$

$\large \int_{0}^{1}\frac{1}{2\sqrt{2}u\sqrt{(u-1/8)^{2}-(3/8)^2}}du$

$\large \int_{0}^{1}\frac{2((u-1/8)+3/8)-((u-1/8)-3/8))}{3\sqrt{2}u\sqrt{(u-1/8)^{2}-(3/8)^2}}du$