How do you find the indefinite integral of $\int x^{2} \sqrt{x^{3} + 3} d x$ $int x^2sqrt(x^3+3)dx$ ?

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Answer 1 · 2016-12-07T10:22:39

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

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

Lets' do this by substitution

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

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

Therefore,

$\int x^{2} \sqrt{x^{3} + 3} d x$ $intx^2sqrt(x^3+3)dx$

$= \frac{1}{3} \int \sqrt{u} d u$ $=1/3intsqrtudu$

$= \frac{1}{3} \int u^{\frac{1}{2}} d u$ $=1/3intu^(1/2)du$

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

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

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

How do you find the indefinite integral of int x^2sqrt(x^3+3)dx∫x2√x3+3dxint x^2sqrt(x^3+3)dx?