How do you integrate $\int 10 x \sqrt{5 x^{2} - 4}$ $int10xsqrt(5x^2-4)$ using substitution?

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Answer 1 · 2016-12-14T06:35:18

The answer is $= \frac{2}{3} {(5 x^{2} - 4)}^{\frac{3}{2}} + C$ $=2/3(5x^2-4)^(3/2)+C$

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

Let $u = 5 x^{2} - 4$ $u=5x^2-4$ , $\Rightarrow$ $=>$ $d u = 10 x d x$ $du=10xdx$

$\int 10 x \sqrt{5 x^{2} - 4} d x = \int \sqrt{u} d u$ $int10xsqrt(5x^2-4)dx=intsqrtudu$

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

$= \frac{2}{3} {(5 x^{2} - 4)}^{\frac{3}{2}} + C$ $=2/3(5x^2-4)^(3/2)+C$

Answer 2 · 2016-12-16T06:20:03

See below, another way to solve this question:
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Continuation:
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How do you integrate int10xsqrt(5x^2-4)∫10x√5x2−4int10xsqrt(5x^2-4) using substitution?