Question

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

Answer 1

`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`

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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`

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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`