1. ## Intrgration by Parts

Hi,

I'm not sure where to begin with find the anti derivative of this function... I know it's IBP, but I'm really confused. Any help is appreciated.

$= \int_ (x^3\sqrt{2+x^2}dx$

2. ## Re: Intrgration by Parts

Do the substitution, let $2+x^2=t$.
Can you go further?

3. ## Re: Intrgration by Parts

If you are using integration by parts then...

$\int x^3\sqrt{2+x^2}dx$

$=\frac{x^4\sqrt{2+x^2}}{4}-\frac{1}{4}\int \frac{x^5}{\sqrt{2+x^2}}dx$

4. ## Re: Intrgration by Parts

Originally Posted by sbhatnagar
If you are using integration by parts then...

$\int x^3\sqrt{2+x^2}dx$

$=\frac{x^4\sqrt{2+x^2}}{4}-\frac{1}{4}\int \frac{x^5}{\sqrt{2+x^2}}dx$
That would be a useful learning experience... but afterwards, integrate the square root instead. Just in case a picture helps...

... where (key in spoiler) ...

Spoiler:

... is the chain rule. Straight continuous lines differentiate downwards (integrate up) with respect to the main variable (in this case x), and the straight dashed line similarly but with respect to the dashed balloon expression (the inner function of the composite which is subject to the chain rule).

But this is wrapped inside the legs-uncrossed version of...

... the product rule, where, again, straight continuous lines are differentiating downwards with respect to x.

... is lazy integration by parts, doing without u and v.

Spoiler:

Spoiler:

__________________________________________________ __________

Don't integrate - balloontegrate!

Balloon Calculus; standard integrals, derivatives and methods

Balloon Calculus Drawing with LaTeX and Asymptote!

5. ## Re: Intrgration by Parts

Hello, sjmiller!

$I \;=\; \int x^3\sqrt{x^2+2}\,dx$

If we must integrate by parts . . .

. . $\begin{Bmatrix} u &=& x^2 && dv &=& x(x^2+2)^{\frac{1}{2}}dx \\ du &=& 2x\,dx ^&& v &=& \frac{1}{3}(x^2+2)^{\frac{3}{2}} \end{Bmatrix}$

$\text{Then: }\:I \;=\;\tfrac{1}{3}x^2(x^2+2)^{\frac{3}{2}} - \tfrac{2}{3}\int x(x^2+2)^{\frac{3}{2}}dx$

. . . . . . . $=\;\tfrac{1}{3}x^2(x^2+2)^{\frac{3}{2}} - \tfrac{2}{15}(x^2+2)^{\frac{5}{2}} + C$

. . . . . . . $=\;\tfrac{1}{15}(x^2+2)^{\frac{3}{2}}\bigg[5x^2 - 2(x^2+2)\bigg] + C$

. . . . . . . $=\;\tfrac{1}{15}(x^2+2)^{\frac{3}{2}}\left(3x^2-4\right) + C$