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

Question

Isabella Ackerman · Accepted Answer

#((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#

Cameron Alexander · Answer

undefined To find the antiderivative of $ \int \frac{x^3}{\sqrt{4x^2 - 1}} \, dx$, you can use the trigonometric substitution method. Let $x = \frac{1}{2}\sec(\theta)$, then $dx = \frac{1}{2}\sec(\theta)\tan(\theta) \, d\theta$. Substituting these into the integral, and simplifying will lead you to the antiderivative.

Samantha Adkison · Answer

undefined To find the antiderivative of ∫ x^3/√(4x^2 - 1) dx, we can use a trigonometric substitution. Let $ x = \frac{1}{2}\sec(\theta) $. Then, $ dx = \frac{1}{2}\sec(\theta)\tan(\theta) d\theta $. Substituting these into the integral, we have:

$ \int \frac{x^3}{\sqrt{4x^2 - 1}} dx = \int \frac{(\frac{1}{2}\sec(\theta))^3}{\sqrt{4(\frac{1}{2}\sec(\theta))^2 - 1}} \cdot \frac{1}{2}\sec(\theta)\tan(\theta) d\theta $

Simplify this expression and integrate it with respect to $ \theta $. Finally, substitute back $ \theta $ in terms of $ x $ to obtain the antiderivative of the original integral.