# Integral with sq. root in it again

1. Nov 27, 2007

### rock.freak667

Integral with sq. root in it....again

1. The problem statement, all variables and given/known data
Find..
$$\int x^\frac{3}{2}\sqrt{1+x} dx$$

2. Relevant equations

3. 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

2. Nov 28, 2007

### Gib Z

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

3. Nov 28, 2007

### rock.freak667

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

4. Nov 28, 2007

### 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.

Last edited: Nov 28, 2007