Integral with sq. root in it again

[rock.freak667]

Integral with sq. root in it...again

Homework Statement

Find..
\int x^\frac{3}{2}\sqrt{1+x} dx

Homework Equations

The Attempt at a Solution

Well I used the fact that:

\sqrt{1+x}=\sum_{n=0} ^\infty \frac{(-1)^n(2n!)x^n}{(1-2n)(n!)^24^n}

and well I just multiplied by x^\frac{3}{2}

so I integrated:
\int \sum_{n=0} ^\infty \frac{(-1)^n(2n!)x^(n+\frac{3}{2}}{(1-2n)(n!)^24^n}

and got \sum_{n=0} ^\infty \frac{(-1)^n(2n!)x^(n+\frac{5}{2}}{(1-2n)(n!)^24^n\frac{5}{2}}

\frac{2x^\frac{5}{2}}{5}\sqrt{1+x} which is wrong because if i differentiate it I get an extra term in it

 

[Gib Z]

That Series is only valid for |x| < 1. I would let x= \sinh^2 u

 

[rock.freak667]

that hyperbolic substitution throws me off as it kinda made it harder for me

 

[Gib Z]

With that substitution I get: 2\int \sinh^4 u \cosh^2 u du, which I believe is possible through methods similar to its circular trigonometric counterpart.

EDIT: I've just done it, its not as easy as I thought but it is possible. Express all the squares in terms of double angle formula. You should get an answer with an x term, and the hyperbolic sines of 2x, 4x and 6x.