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#