How do you integrate $x^{5} \cdot \sqrt{x^{3} + 1} d x$ $x^5 * sqrt(x^3+1) dx$ ?

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How do you integrate $x^{5} \cdot \sqrt{x^{3} + 1} d x$ $x^5 * sqrt(x^3+1) dx$ ?

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Answer 1 · 2017-05-02T16:04:56

The integral is $\frac{2}{15} {(x^{3} + 1)}^{\frac{5}{2}} - \frac{2}{9} {(x^{3} + 1)}^{\frac{3}{2}} + C$ $2/15(x^3 + 1)^(5/2) - 2/9(x^3 + 1)^(3/2) + C$

Explanation:

Let $u = x^{3} + 1$ $u = x^3 + 1$ . Then $d u = 3 x^{2} d x \to d x = \frac{d u}{3 x^{2}}$ $du = 3x^2dx -> dx =(du)/(3x^2)$ . Let $I$ $I$ be the integral.

$I = \int x^{5} \sqrt{u} \cdot \frac{d u}{3 x^{2}}$ $I= intx^5sqrt(u) * (du)/(3x^2)$

$I = \frac{1}{3} \int x^{3} \sqrt{u} d u$ $I = 1/3intx^3sqrt(u)du$

$I = \frac{1}{3} \int (u - 1) \sqrt{u} d u$ $I = 1/3int(u - 1)sqrt(u)du$

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

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

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

$I = \frac{2}{5} (\frac{1}{3}) u^{\frac{5}{2}} - \frac{2}{3} (\frac{1}{3}) u^{\frac{3}{2}} + C$ $I = 2/5(1/3)u^(5/2) - 2/3(1/3)u^(3/2) + C$

$I = \frac{2}{15} u^{\frac{5}{2}} - \frac{2}{9} u^{\frac{3}{2}} + C$ $I = 2/15u^(5/2) - 2/9u^(3/2) +C$

$I = \frac{2}{15} {(x^{3} + 1)}^{\frac{5}{2}} - \frac{2}{9} {(x^{3} + 1)}^{\frac{3}{2}} + C$ $I = 2/15(x^3 + 1)^(5/2) - 2/9(x^3 + 1)^(3/2) + C$

Hopefully this helps!

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How do you integrate $x^{5} \cdot \sqrt{x^{3} + 1} d x$ $x^5 * sqrt(x^3+1) dx$ ?

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