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

Question

Aurora Alvarado · Accepted Answer

The answer is #=xln(1+x^2)-2x+2arctanx+C#

Utilizing integration by parts, determine this integral.

#intuv'=uv-intu'v#

#u=ln(1+x^2)#, #=>#, #u'=(2x)/(1+x^2)#

#v'=1#, #=>#, #v=x#

Consequently, the integral is

#intln(1+x^2)dx=xln(1+x^2)-int(2x^2dx)/(1+x^2)#

The subsequent integral is

#int(2x^2dx)/(1+x^2)=2int((x^2+1-1)dx)/(1+x^2)#

#=2int1dx-2int(1dx)/(1+x^2)#

#=2x-2arctanx#

Ultimately, the integral is

#I=xln(1+x^2)-2x+2arctanx+C#

Christopher Allers · Answer

undefined To find the integral of ln(1+x^2), you can use the substitution method. Let $ u = 1 + x^2 $. Then, $ du/dx = 2x $, and $ dx = du/(2x) $. Substitute these into the integral to get:

$$ \int \ln(1+x^2) \, dx = \int \ln(u) \frac{du}{2x} $$

Using integration by parts, let $ dv = du/(2x) $ and $ u = \ln(u) $. Then, $ v = u \ln(u) - \int u \, du/u $.

$$ \int \ln(1+x^2) \, dx = x \ln(1+x^2) - \frac{1}{2} \int \frac{2x^2 \, dx}{1+x^2} $$

Now, use a trigonometric substitution, letting $ x = \tan(\theta) $. Then, $ dx = \sec^2(\theta) \, d\theta $.

$$ \int \ln(1+x^2) \, dx = x \ln(1+x^2) - \frac{1}{2} \int \frac{2\tan^2(\theta) \sec^2(\theta) \, d\theta}{1+\tan^2(\theta)} $$

$$ = x \ln(1+x^2) - \frac{1}{2} \int \sin^2(\theta) \, d\theta $$

$$ = x \ln(1+x^2) - \frac{1}{4} (\theta - \sin(\theta) \cos(\theta)) + C $$

$$ = x \ln(1+x^2) - \frac{1}{4} (\arctan(x) - x/(1+x^2)) + C $$

where $ C $ is the constant of integration.