integrate x^3sqrt(1+x^2)dx

[Amandar]

K so I've done this integral soo many times, and I can't get the right answer.. the integral is x^3sqrt(1+x^2)dx , and the answer is supposed to be 1/15(x^2+1)^3/2(3x^2 -2) .. but i can't seem to do it. I've tried doing substitution using u=x^2 which didn't work. and than i thought about using 1+x^2 as u, but there is no 1+ after the intergral in order to get du... thanks in advance for any help

 

[stapel]

I've done this integral soo many times....

What did you try? What results did you get?

Please be complete. Thank you!

Eliz.

 

[Amandar]

I've tried doing substitution using u=x^2 which didn't work. and than i thought about using 1+x^2 as u, but there is no 1+ after the intergral in order to get du

when i tried with u=x^2 i got .. 1/2Susqrt(u+1)du, than i let v = u+1 and ended up with 1/2*x^4/2*-2/3(x^2+1)^3/2

 

[PAULK]

Try a 'rationalizing substitution':

Let u = sqrt(1 + x^2)

u^2 = 1 + x^2

u^2 - 1 = x^2

2u du = 2x dx
u du = x dx
Now work on the integral:
{
| x^3 sqrt(1 + x^2) dx =
}

{
| x^2 sqrt(1 + x^2) x dx
}

{
| (u^2 - 1) u u du
}

You can do the rest, I think.

 

[Amandar]

so this is what i've gotten now.....

\(\displaystyle \int {x^3 \sqrt {x^2 + 1} dx}\)
\(\displaystyle u = x^2 + 1\)
\(\displaystyle du = 2xd2\)
\(\displaystyle x^2 = u - 1\)
\(\displaystyle = \frac{1}{2}\int {x^2 \sqrt {x^2 + 1} 2xdx}\)
\(\displaystyle = \frac{1}{2}\int {x^2 \sqrt u du}\)
\(\displaystyle = \frac{1}{2}\int {(u - 1)\sqrt u du}\)
\(\displaystyle = \frac{1}{2}(\frac{{u^2 }}{2} - u)\frac{2}{3}{u^\frac{3}{2}\)
\(\displaystyle = \frac{1}{6}((x^2 + 1)^2 - (x^2 + 1))(x^2 - 1)^\frac{3}{2}\)
\(\displaystyle = \frac{1}{6}(x^4 + 2x^2 + 1 - x^2 + 1)(x^2 + 1)^\frac{3}{2}\)
\(\displaystyle = \frac{1}{6}(x^4 + x^2 + 2)(x^2 + 1)^\frac{3}{2}\)

but the answer is supposed to be

\(\displaystyle \frac{1}{15}(x^2 + 1)^\frac{3}{2} (3x^2 - 2)\)


any idea where im going wrong?? :|

 

[stapel]

I'm sorry, but I don't follow your integration...?

Shouldn't u[sup:2k2itios]3/2[/sup:2k2itios] integrate as (2/5)u[sup:2k2itios]5/2[/sup:2k2itios], and u[sup:2k2itios]1/2[/sup:2k2itios] integrate as (2/3)u[sup:2k2itios]3/2[/sup:2k2itios]...? :?:

Thank you!

Eliz.