What is $int 4xln(x^2+1) dx$ ?

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Answer 1 · 2018-04-04T09:27:05

$int 4xln(x^2+1)dx = 2 (x^2+1)ln(x^2+1)-2x^2 +C$

Substitute:

$u = x^2+1$

$du = 2xdx$

so:

$int 4xln(x^2+1)dx = 2 int ln u du$

Integrate now by parts:

$int ln u du = u ln u - int u d(lnu)$

$int ln u du = u ln u - int u (du)/u$

$int ln u du = u ln u - int du$

$int ln u du = u ln u -u +C$

$int ln u du = u (ln u -1) +C$

undoing the substitution:

$int 4xln(x^2+1)dx = 2 (x^2+1)(ln(x^2+1)-1) +C$

and simplifying:

$int 4xln(x^2+1)dx = 2 (x^2+1)ln(x^2+1)-2x^2 +C$

where we absorbed all constants in $C$ .

What is int 4xln(x^2+1) dxint 4xln(x^2+1) dx?