How do you integrate $(- \frac{3}{x}) + (\frac{ln x}{x^{3}}) d x$ $(-3/x) + (lnx/x^3) dx$ ?

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Answer 1 · 2018-03-23T05:20:29

$- 3 ln | x | - \frac{1}{4 x^{2}} (2 ln x + 1) + C, or,$ $-3ln|x|-1/(4x^2)(2lnx+1)+C, or,$

$- 3 ln | x | - \frac{1}{4 x^{2}} ln (e x^{2}) + C$ $-3ln|x|-1/(4x^2)ln(ex^2)+C$ .

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

$\int u v d x = u \int v d x - \int {\frac{d u}{d x} \int v d x} d x$ $intuvdx=uintvdx-int{(du)/dxintvdx}dx$ .

Let, $I = \int (- \frac{3}{x} + \frac{ln x}{x^{3}}) d x$ $I=int(-3/x+lnx/x^3)dx$ ,

$= - 3 \int \frac{1}{x} d x + \int \frac{ln x}{x^{3}} d x$ $=-3int1/xdx+intlnx/x^3dx$ ,

$= - 3 ln | x | + I_{1}, where, I_{1} = \int \frac{ln x}{x^{3}} d x$ $=-3ln|x|+I_1," where, "I_1=intlnx/x^3dx$ .

For $I_{1}$ $I_1$ , we use IBP with, $u = ln x, and, v = \frac{1}{x^{3}} = x^{- 3}$ $u=lnx, and, v=1/x^3=x^-3$ .

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

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

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

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

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

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

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

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How do you integrate (-3/x) + (lnx/x^3) dx(−3x)+(lnxx3)dx(-3/x) + (lnx/x^3) dx?