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.
$\displaystyle = \int_ (x^3\sqrt{2+x^2}dx$
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:
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The sub instead...
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Balloon Calculus; standard integrals, derivatives and methods
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Hello, sjmiller!
$\displaystyle I \;=\; \int x^3\sqrt{x^2+2}\,dx$
If we must integrate by parts . . .
. . $\displaystyle \begin{Bmatrix} u &=& x^2 && dv &=& x(x^2+2)^{\frac{1}{2}}dx \\ du &=& 2x\,dx ^&& v &=& \frac{1}{3}(x^2+2)^{\frac{3}{2}} \end{Bmatrix}$
$\displaystyle \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 $
. . . . . . .$\displaystyle =\;\tfrac{1}{3}x^2(x^2+2)^{\frac{3}{2}} - \tfrac{2}{15}(x^2+2)^{\frac{5}{2}} + C $
. . . . . . .$\displaystyle =\;\tfrac{1}{15}(x^2+2)^{\frac{3}{2}}\bigg[5x^2 - 2(x^2+2)\bigg] + C$
. . . . . . .$\displaystyle =\;\tfrac{1}{15}(x^2+2)^{\frac{3}{2}}\left(3x^2-4\right) + C$