integrating help

**stud_02** · April 30th 2009, 08:38 AM

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

with substitution

**redsoxfan325** · April 30th 2009, 12:57 PM

Originally Posted by stud_02

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

with substitution

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

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

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

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

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:

**simplependulum** · May 1st 2009, 01:58 AM

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

**adhyeta** · May 1st 2009, 09:42 AM

starting from redsox's...

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

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

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

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

that should help you.

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