How do you integrate #(-3/x) + (lnx/x^3) dx#?

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Answer 1 · 2018-03-23T05:20:29

# -3ln|x|-1/(4x^2)(2lnx+1)+C, or, #

#-3ln|x|-1/(4x^2)ln(ex^2)+C#.

Prerequisite : The Rule of Integration by Parts (IBP) :

#intuvdx=uintvdx-int{(du)/dxintvdx}dx#.

Let, #I=int(-3/x+lnx/x^3)dx#,

#=-3int1/xdx+intlnx/x^3dx#,

#=-3ln|x|+I_1," where, "I_1=intlnx/x^3dx#.

For #I_1#, we use IBP with, #u=lnx, and, v=1/x^3=x^-3#.

#:. (du)/dx=1/x, and, intvdx=x^(-3+1)/(-3+1)=-1/(2x^2)#.

#:. I_1=-lnx/(2x^2)-int{(1/x)(-1/(2x^2))}dx#,

#=-lnx/(2x^2)+1/2int1/x^3dx#,

#=-lnx/(2x^2)+1/2*x^(-3+1)/(-3+1)#,

#=-lnx/(2x^2)-1/(4x^2)#.

# rArr I=-3ln|x|-1/(4x^2)(2lnx+1)+C, or, #

# I=-3ln|x|-1/(4x^2)ln(ex^2)+C#.

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