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

Question

Aurora Alvarado · Accepted Answer

#I=intx^3/(4+x^2)dx# Let #u=4+x^2#, implying that #du=2xdx#. Also note that #x^2=u-4#. Rearranging the integral: #I=1/2int(x^2(2xdx))/(4+x^2)=1/2int(u-4)/udu=1/2int(1-4/u)du# Splitting up the integral: #I=1/2intdu-2int1/udu=1/2u-2lnabsu=1/2(4+x^2)-2lnabs(4+x^2)+C# Letting the constant from #1/2(4+x^2)# absorb into #C#: #I=1/2x^2-2ln(x^2+4)+C#

Emilia Aguilar · Answer

undefined To find the antiderivative of $ \int \frac{x^3}{4+x^2} \, dx $, you can use substitution. Let $ u = 4 + x^2 $. Then $ du = 2x \, dx $. Rearranging, $ \frac{1}{2} du = x \, dx $. Now substitute these into the integral:

$ \int \frac{x^3}{4+x^2} \, dx = \int \frac{x^2 \cdot x}{4+x^2} \, dx $

Substitute $ u = 4 + x^2 $ and $ \frac{1}{2} du = x \, dx $:

$ = \frac{1}{2} \int \frac{x^2}{u} \, du $

Now, we can split $ \frac{x^2}{u} $ into $ \frac{x^2}{4} \cdot \frac{1}{u} $:

$ = \frac{1}{2} \cdot \frac{1}{4} \int \frac{x^2}{\frac{1}{4}(4+x^2)} \, du $

Now, we have a simpler integral:

$ = \frac{1}{8} \int \frac{1}{u} \, du $

Integrate $ \frac{1}{u} $:

$ = \frac{1}{8} \ln|u| + C $

Finally, substitute back $ u = 4 + x^2 $:

$ = \frac{1}{8} \ln|4 + x^2| + C $

So, the antiderivative of $ \int \frac{x^3}{4+x^2} \, dx $ is $ \frac{1}{8} \ln|4 + x^2| + C $, where $ C $ is the constant of integration.