Integration problem?

Question

integral {ln(1+x^2)} dx

Answer 1

Integral (ln (1 + x^2) dx)

Let's do this by parts.

Let u = ln(1 + x^2). dv = dx
du = 2x/(1 + x^2) . v = x

Therefore, this becomes

x ln(1 + x^2) - Integral (2x^2 / (1 + x^2) ) dx

Pulling the constant 2 out,

x ln(1 + x^2) - 2 * Integral (x^2 / (1 + x^2) dx )

Long division on (x^2 + 1) into (x^2) will allow us to change
x^2 / (1 + x^2) into 1 - 1/(1 + x^2), so we have

x ln(1 + x^2) - 2 * Integral [1 - 1/(1 + x^2)] dx

Which we can now integrate normally, keeping in mind that
1/(1 + x^2) is one of our known derivatives (derivative of arctan(x)).

x ln(1 + x^2) - 2[x - arctan(x)] + C

x ln(1 + x^2) - 2x + 2arctan(x) + C

Answer 2

integral {log(1+x^2)} dx
Let us suppose the above equation as:

integral{1*[log(1+x^2)]} dx
integrating by parts, we get

=> {log(1+x^2)} * 1dx - [d/dx{log(1+x^2)} * 1dx]
=> {log(1+x^2)} * x - [ {1/(1+x^2)}(2x) * x]
=> x{log(1+x^2)} - 2x^2/(1+x^2)