How do you integrate Ln(1+x^2)?

1 Answer
sente
Jul 17, 2016

intln(1+x^2)dx=xln(1+x^2)-2x+arctan(x)+C

Explanation:

First, applying integration by parts, we let

u = ln(1+x^2) and dv = dx
=> du = (2x)/(1+x^2) and v = x

Applying the formula intudv = uv-intvdu, we have

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

To solve the remaining integral, we will use trig substitution.

Let x = tan(theta)
=> dx = sec^2(theta)d theta and theta = arctan(x)

Substituting, we have

intx^2/(1+x^2)dx = inttan^2(theta)/(1+tan^2(theta))sec^2(theta)d theta

=inttan^2(theta)/sec^2(theta)sec^2(theta)d theta

=inttan^2(theta)d theta

=int(sec^2(theta)-1)d theta

=intsec^2(theta)d theta - intd theta

=tan(theta)-theta + C

=x - arctan(x) + C

Going back to our original problem, we have

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

=xln(1+x^2)-2(x-arctan(x))+C

=xln(1+x^2)-2x+arctan(x)+C

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