How do you integrate $int10xsqrt(5x^2-4)$ using substitution?

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Answer 1 · 2016-12-14T06:35:18

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

We use, $intx^ndx=x^(n+1)/(n+1)+C$ $(n!=-1)$

Let $u=5x^2-4$ , $=>$ $du=10xdx$

$int10xsqrt(5x^2-4)dx=intsqrtudu$

$=intu^(1/2)du=u^(3/2)/(3/2)$

$=2/3(5x^2-4)^(3/2)+C$

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)int10xsqrt(5x^2-4) using substitution?