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#