How do you find the integral of # x^3/(1+x^2)#?

Question

Ezra Adams · Accepted Answer

#1/2(1+x^2-ln(1+x^2)) + C# We'll do integration by substitution. Take: #u = x^2# #du = 2x# Substituting this in, we get: #1/2\intu/(1+u)du# Now take: #t = 1 + u# #dt = du# Substitute this in: #1/2\int(t-1)/tdt = 1/2\int1 - 1/tdt = 1/2(\int1dt - \int 1/tdt)# #=1/2(t - ln(t)) + C# Now we need to change back to #x#: #=1/2(1+u-ln(1+u)) + C# #=1/2(1+x^2-ln(1+x^2)) + C#

Cameron Alexander · Answer

undefined To find the integral of $ \frac{x^3}{1+x^2} $, you can use a substitution method. Let $ u = 1 + x^2 $. Then $ du = 2x \, dx $. Rearranging for $ dx $, we get $ \frac{du}{2x} = dx $.

Substituting $ u $ and $ dx $ into the integral:

$$ \int \frac{x^3}{1+x^2} \, dx = \int \frac{x^3}{u} \cdot \frac{du}{2x} $$

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

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

Now, use another substitution: $ v = 1 + x^2 $, so $ dv = 2x \, dx $. This time we need to express $ x^2 $ in terms of $ v $. Since $ v = 1 + x^2 $, $ x^2 = v - 1 $.

Substituting $ v $ and $ x^2 $ into the integral:

$$ \frac{1}{2} \int \frac{x^2}{1+x^2} \, du = \frac{1}{2} \int \frac{v - 1}{v} \, dv $$

$$ = \frac{1}{2} \int \left( 1 - \frac{1}{v} \right) \, dv $$

$$ = \frac{1}{2} \left( v - \ln|v| \right) + C $$

Substitute back $ v = 1 + x^2 $ to get the final result:

$$ = \frac{1}{2} \left( 1 + x^2 - \ln|1+x^2| \right) + C $$

So, the integral of $ \frac{x^3}{1+x^2} $ is $ \frac{1}{2} \left( 1 + x^2 - \ln|1+x^2| \right) + C $, where $ C $ is the constant of integration.