How do you find the integral $\int x^{2} \sqrt{x^{3} + 1} d x$ $int x^2sqrt(x^3+1)dx$ using substitution from [-1,1]?

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How do you find the integral $\int x^{2} \sqrt{x^{3} + 1} d x$ $int x^2sqrt(x^3+1)dx$ using substitution from [-1,1]?

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Answer 1 · 2016-12-27T15:32:38

The answer is $= \frac{4}{9} \sqrt{2}$ $=4/9sqrt2$

Explanation:

We need

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

We do a substitution

Let $u = x^{3} + 1$ $u=x^3+1$ , $\Rightarrow$ $=>$ , $d u = 3 x^{2} d x$ $du=3x^2dx$

Therefore,

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

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

So,

$\int_{- 1}^{1} x^{2} \sqrt{x^{3} + 1} d x = {[\frac{2}{9} {(x^{3} + 1)}^{\frac{3}{2}}]}_{- 1}^{\frac{3}{2}}$ $int_-1 ^1 x^2sqrt(x^3+1)dx= [2/9(x^3+1)^(3/2)]_ -1^1$

$= \frac{4}{9} \sqrt{2} - 0 = \frac{4}{9} \sqrt{2}$ $=4/9sqrt2-0=4/9sqrt2$

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How do you find the integral $\int x^{2} \sqrt{x^{3} + 1} d x$ $int x^2sqrt(x^3+1)dx$ using substitution from [-1,1]?

1 Answer

Explanation:

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How do you find the integral ∫x2√x3+1dxint x^2sqrt(x^3+1)dx using substitution from [-1,1]?

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How do you find the integral $\int x^{2} \sqrt{x^{3} + 1} d x$ $int x^2sqrt(x^3+1)dx$ using substitution from [-1,1]?