How do you find the antiderivative of $int x^3/sqrt(4x^2-1)dx$ ?

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Answer 1 · 2016-11-05T22:19:51

$((2x^2+1)sqrt(4x^2-1))/24+C$

$I=intx^3/sqrt(4x^2-1)dx$

Let $u=4x^2-1$ so that $du=8xdx$ . Also note that $x^2=(u+1)/4$ , which will be useful in a second:

$I=1/8int(x^2(8x))/sqrt(4x^2-1)dx$

Substituting in our $8xdx$ and $x^2$ terms we have:

$I=1/8int((u+1)/4)/sqrtudu=1/32int(u+1)/sqrtudu$

$I=1/32int(u^(1/2)+u^(-1/2))du$

Now integrating using the power rule for integration:

$I=1/32(u^(3/2)/(3/2)+u^(1/2)/(1/2))=1/32(2/3u^(3/2)+2u^(1/2))$

$I=1/48u^(3/2)+1/16u^(1/2)=(u^(3/2)+3u^(1/2))/48=(u^(1/2)(u+3))/48$

Now since $u=4x^2-1$ :

$I=(sqrt(4x^2-1)(4x^2-1+3))/48=(sqrt(4x^2-1)(4x^2+2))/48$

So:

$I=(sqrt(4x^2-1)(2x^2+1))/24+C$

How do you find the antiderivative of int x^3/sqrt(4x^2-1)dxint x^3/sqrt(4x^2-1)dx?