How do you integrate $\int x^{3} \cdot \frac{d x}{{(x^{2} - 1)}^{\frac{2}{3}}}$ $int x^3*dx/(x^2-1)^(2/3)$ ?

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How do you integrate $\int x^{3} \cdot \frac{d x}{{(x^{2} - 1)}^{\frac{2}{3}}}$ $int x^3*dx/(x^2-1)^(2/3)$ ?

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Answer 1 · 2016-08-08T16:45:10

$\int \frac{x^{3}}{{(x^{2} - 1)}^{\frac{2}{3}}} d x = \frac{3 {(x^{2} - 1)}^{\frac{1}{3}} (x^{2} + 3)}{8} + C$ $intx^3/(x^2-1)^(2/3)dx=(3(x^2-1)^(1/3)(x^2+3))/8+C$

Explanation:

We have the integral:

$I = \int \frac{x^{3}}{{(x^{2} - 1)}^{\frac{2}{3}}} d x$ $I=intx^3/(x^2-1)^(2/3)dx$

Using substitution, let $u = x^{2} - 1$ $u=x^2-1$ , so $d u = 2 x d x$ $du=2xdx$ and $x^{2} = u + 1$ $x^2=u+1$ . Thus:

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

Split up the fraction and then the integrals:

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

Integrate using the power rule for integrals:

$I = \frac{1}{2} (\frac{u^{\frac{4}{3}}}{\frac{4}{3}}) + \frac{1}{2} (\frac{u^{\frac{1}{3}}}{\frac{1}{3}}) = \frac{3}{8} u^{\frac{4}{3}} + \frac{3}{2} u^{\frac{1}{3}} = \frac{3 u^{\frac{4}{3}} + 12 u^{\frac{1}{3}}}{8}$ $I=1/2(u^(4/3)/(4/3))+1/2(u^(1/3)/(1/3))=3/8u^(4/3)+3/2u^(1/3)=(3u^(4/3)+12u^(1/3))/8$

Continuing simplification:

$I = \frac{3 u^{\frac{1}{3}} (u + 4)}{8} = \frac{3 {(x^{2} - 1)}^{\frac{1}{3}} ((x^{2} - 1) + 4)}{8} = \frac{3 {(x^{2} - 1)}^{\frac{1}{3}} (x^{2} + 3)}{8} + C$ $I=(3u^(1/3)(u+4))/8=(3(x^2-1)^(1/3)((x^2-1)+4))/8=(3(x^2-1)^(1/3)(x^2+3))/8+C$

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How do you integrate $\int x^{3} \cdot \frac{d x}{{(x^{2} - 1)}^{\frac{2}{3}}}$ $int x^3*dx/(x^2-1)^(2/3)$ ?

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How do you integrate $\int x^{3} \cdot \frac{d x}{{(x^{2} - 1)}^{\frac{2}{3}}}$ $int x^3*dx/(x^2-1)^(2/3)$ ?