How do you solve this integral?

Question

$\int x^{3} {(x^{2} - 4)}^{\frac{3}{2}}$ $intx^3(x^2-4)^(3/2)$

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Answer 1 · 2018-02-16T08:18:48

$I = \frac{1}{7} {(x^{2} - 4)}^{\frac{7}{2}} + \frac{4}{5} {(x^{2} - 4)}^{\frac{5}{2}} + C$ $I=1/7(x^2-4)^(7/2)+4/5(x^2-4)^(5/2)+C$

We want to solve

$I = \int x^{3} {(x^{2} - 4)}^{\frac{3}{2}} d x$ $I=intx^3(x^2-4)^(3/2)dx$

Use substitution let $u = x^{2} - 4 \Rightarrow \frac{d u}{d x} = 2 x$ $u=x^2-4=>(du)/dx=2x$

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

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

But $u = x^{2} - 4 \Leftrightarrow x^{2} = u + 4$ $u=x^2-4<=>x^2=u+4$

$I = \frac{1}{2} \int (u + 4) {(u)}^{\frac{3}{2}} d u$ $I=1/2int(u+4)(u)^(3/2)du$

$= \frac{1}{2} \int u {(u)}^{\frac{3}{2}} d u + \frac{1}{2} \int 4 {(u)}^{\frac{3}{2}} d u$ $=1/2intu(u)^(3/2)du+1/2int4(u)^(3/2)du$

$= \frac{1}{2} \int u^{\frac{5}{2}} d u + 2 \int u^{\frac{3}{2}} d u$ $=1/2intu^(5/2)du+2intu^(3/2)du$

By the power rule for the integrals

$I = \frac{1}{7} u^{\frac{7}{2}} + \frac{4}{5} u^{\frac{5}{2}} + C$ $I=1/7u^(7/2)+4/5u^(5/2)+C$

Substitute $u = x^{2} - 4$ $u=x^2-4$ back

$I = \frac{1}{7} {(x^{2} - 4)}^{\frac{7}{2}} + \frac{4}{5} {(x^{2} - 4)}^{\frac{5}{2}} + C$ $I=1/7(x^2-4)^(7/2)+4/5(x^2-4)^(5/2)+C$