How do you integrate $int ln x^2 dx$ using integration by parts?

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Answer 1 · 2016-01-17T13:05:43

$int ln x^2 dx=x* ln x^2 - 2x + C$

The formula $int u *dv= u*v-int v* du$

the given $int ln x^2 dx$
Let $u=ln x^2$
$dv =dx$

$v=x$

$du=2x*dx/x^2=2* dx/x$

$int ln x^2 dx=x*ln x^2 - int x*2*dx/x+C$

$int ln x^2 dx=x*ln x^2 - int 2*dx+C$

$int ln x^2 dx=x* ln x^2 - 2x + C$

How do you integrate int ln x^2 dx int ln x^2 dx using integration by parts?