Question

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

Answer 1

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

Explanation:

`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`