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What is the integral of #(x^2)(lnx)#?

1 Answer

Apr 6, 2018

#int x^2*Lnx*dx=x^3/3*Lnx-x^3/9+C#

Explanation:

After setting #dv=x^2*dx# and #u=Lnx# for using integration by parts, #v=x^3/3# and #du=dx/x#

Hence,

#int udv=uv-int vdu#

#int x^2*Lnx*dx=x^3/3*Lnx-int x^3/3*dx/x#

=#x^3/3*Lnx-int x^2/3*dx#

=#x^3/3*Lnx-x^3/9+C#

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