FIND #int_. (x^2)ln 5x dx# using IBP (Integration by parts) ?

1 Answer
Øko
Mar 8, 2018

#I=1/9x^3(3ln(5x)-1)+C#

Explanation:

We want to integrate

#I=intx^2ln(5x)dx#

Use integration by parts

#intudv=uv-intvdu#

Let #u=ln(5x)# and #dv=x^2dx#

Then #du=1/xdx# and #v=1/3x^3#

Thus

#I=ln(5x)1/3x^3-int1/3x^3*1/xdx#

#color(white)(I)=1/3ln(5x)x^3-1/3intx^2dx#

#color(white)(I)=1/3ln(5x)x^3-1/9x^3+C#

#color(white)(I)=1/9x^3(3ln(5x)-1)+C#

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