How do you integrate $x^3 sqrt(x^2 + 1) dx$ ?

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Answer 1 · 2018-04-25T15:53:43

The answer $(x^2+1)^(5/2)/5-(x^2+1)^(3/2)/3+c$

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$intx^3*(x^2+1)^(1/2)*dx$

suppose:
$u=x^2+1$
$x=(u-1)^(1/2)$
$du=2x*dx$
$dx=1/(2x)*du$
the int become after suppose:

$int[sqrt(u-1)]^3*sqrt(u)*1/(2*sqrt(u-1))*du$

$1/2int[sqrt(u-1)]^2*sqrtu*du$

$1/2int(u-1)*sqrtu*du=1/2intu*sqrtu-sqrtu*du=1/2intu^(3/2)-u^(1/2)*du$

$1/2[2/5*u^(5/2)-2/3*u^(3/2)]+c$

$u^(5/2)/5-u^(3/2)/3+c$

$(x^2+1)^(5/2)/5-(x^2+1)^(3/2)/3+c$

How do you integrate x^3 sqrt(x^2 + 1) dxx^3 sqrt(x^2 + 1) dx?