∫ (ln( x^2 + 1 )
Think of it as the product 1*ln(x^2 + 1)
Then use integration by parts with
u = ln(x^2 + 1), du = 2x/(x^2 + 1), dv = 1, v = x
When these are fitted into the formula
INT (u*dv) dx = u*v - INT (v*du) dx
it turns it into something much easier to integrate.
Have a go at it. Post additional note if you want more hints.
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Verified answer
Think of it as the product 1*ln(x^2 + 1)
Then use integration by parts with
u = ln(x^2 + 1), du = 2x/(x^2 + 1), dv = 1, v = x
When these are fitted into the formula
INT (u*dv) dx = u*v - INT (v*du) dx
it turns it into something much easier to integrate.
Have a go at it. Post additional note if you want more hints.