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

Mar 23, 2018

$- 3 \ln | x | - \frac{1}{4 {x}^{2}} \left(2 \ln x + 1\right) + C , \mathmr{and} ,$

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

#### Explanation:

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

$\int u v \mathrm{dx} = u \int v \mathrm{dx} - \int \left\{\frac{\mathrm{du}}{\mathrm{dx}} \int v \mathrm{dx}\right\} \mathrm{dx}$.

Let, $I = \int \left(- \frac{3}{x} + \ln \frac{x}{x} ^ 3\right) \mathrm{dx}$,

$= - 3 \int \frac{1}{x} \mathrm{dx} + \int \ln \frac{x}{x} ^ 3 \mathrm{dx}$,

$= - 3 \ln | x | + {I}_{1} , \text{ where, } {I}_{1} = \int \ln \frac{x}{x} ^ 3 \mathrm{dx}$.

For ${I}_{1}$, we use IBP with, $u = \ln x , \mathmr{and} , v = \frac{1}{x} ^ 3 = {x}^{-} 3$.

$\therefore \frac{\mathrm{du}}{\mathrm{dx}} = \frac{1}{x} , \mathmr{and} , \int v \mathrm{dx} = {x}^{- 3 + 1} / \left(- 3 + 1\right) = - \frac{1}{2 {x}^{2}}$.

$\therefore {I}_{1} = - \ln \frac{x}{2 {x}^{2}} - \int \left\{\left(\frac{1}{x}\right) \left(- \frac{1}{2 {x}^{2}}\right)\right\} \mathrm{dx}$,

$= - \ln \frac{x}{2 {x}^{2}} + \frac{1}{2} \int \frac{1}{x} ^ 3 \mathrm{dx}$,

$= - \ln \frac{x}{2 {x}^{2}} + \frac{1}{2} \cdot {x}^{- 3 + 1} / \left(- 3 + 1\right)$,

$= - \ln \frac{x}{2 {x}^{2}} - \frac{1}{4 {x}^{2}}$.

$\Rightarrow I = - 3 \ln | x | - \frac{1}{4 {x}^{2}} \left(2 \ln x + 1\right) + C , \mathmr{and} ,$

$I = - 3 \ln | x | - \frac{1}{4 {x}^{2}} \ln \left(e {x}^{2}\right) + C$.

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