Integrate $intx^3/sqrt(x^2+4)$ using trig substitution?

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Answer 1 · 2017-10-11T06:43:38

See the explanatiom below

You have to change as follows

$I=8(1/3u^3-u)$

$I=8/3(sec^3theta-3sectheta)$

$=8/3(((x^2+1)/2)^(3/2)-3sec(arctan(x/2))+C$

It's easier without trigonometric substitution

Let $u=x^2+4$ , $=>$ , $du=2xdx$

$I=1/2int((u-4)du)/sqrtu$

$=1/2intsqrtudu-int4/sqrtudu$

$=(u^(3/2)/3-4sqrtu)$

$=1/3(x^2+4)^(3/2)-4sqrt(x^2+4)$

$=((x^2-8))/3sqrt(x^2+4)+C$

Integrate intx^3/sqrt(x^2+4)intx^3/sqrt(x^2+4) using trig substitution?