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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