How do you integrate $\int x^{2} \sqrt{x^{3} + 1}$ $int x^2sqrt(x^3+1)$ using substitution?

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Answer 1 · 2017-06-25T19:39:42

The answer is $= \frac{2}{9} {(x^{3} + 1)}^{\frac{3}{2}} + C$ $=2/9(x^3+1)^(3/2)+C$

We need

$\int x^{n} d x = \frac{x^{n + 1}}{n + 1} + C (n \neq - 1)$ $intx^ndx=x^(n+1)/(n+1)+C(n!=-1)$

Perform the substitution

$u = x^{3} + 1$ $u=x^3+1$

$d u = 3 x^{2} d x$ $du=3x^2dx$

$x^{2} d x = \frac{d u}{3}$ $x^2dx=(du)/3$

Therefore,

$\int x^{2} \sqrt{x^{3} + 1} d x = \frac{1}{3} \int u^{\frac{1}{2}} d u$ $intx^2sqrt(x^3+1)dx=1/3intu^(1/2)du$

$= \frac{1}{3} \frac{u^{\frac{3}{2}}}{\frac{3}{2}}$ $=1/3u^(3/2)/(3/2)$

$= \frac{2}{9} u^{\frac{3}{2}}$ $=2/9u^(3/2)$

$= \frac{2}{9} {(x^{3} + 1)}^{\frac{3}{2}} + C$ $=2/9(x^3+1)^(3/2)+C$

How do you integrate ∫x2√x3+1int x^2sqrt(x^3+1) using substitution?