# Integrate 1/(x^2)ln(x)

## \int\frac{1}{x^2}\cdot\ln\left(x\right)dx

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$-\frac{1}{x}-\frac{1}{x}\ln\left(x\right)+C_0$

## Step by step solution

Problem

$\int\frac{1}{x^2}\cdot\ln\left(x\right)dx$
1

Use the integration by parts theorem to calculate the integral $\int\frac{1}{x^2}\ln\left(x\right)dx$, using the following formula

$\displaystyle\int u\cdot dv=u\cdot v-\int v \cdot du$
2

First, identify $u$ and calculate $du$

$\begin{matrix}\displaystyle{u=\ln\left(x\right)}\\ \displaystyle{du=\frac{1}{x}dx}\end{matrix}$
3

Now, identify $dv$ and calculate $v$

$\begin{matrix}\displaystyle{dv=\frac{1}{x^2}dx}\\ \displaystyle{\int dv=\int \frac{1}{x^2}dx}\end{matrix}$
4

Solve the integral

$v=\int\frac{1}{x^2}dx$
5

Rewrite the exponent using the power rule $\frac{a^m}{a^n}=a^{m-n}$, where in this case $m=0$

$\int x^{-2}dx$
6

Apply the power rule for integration, $\displaystyle\int x^n dx=\frac{x^{n+1}}{n+1}$, where $n$ represents a constant function

$\frac{x^{-1}}{-1}$
7

Now replace the values of $u$, $du$ and $v$ in the last formula

$\int\frac{1}{x}x^{-1}dx-x^{-1}\ln\left(x\right)$
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Multiplying the fraction and term

$\int\frac{x^{-1}}{x}dx-x^{-1}\ln\left(x\right)$
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Simplifying the fraction by $x$

$\int x^{-2}dx-x^{-1}\ln\left(x\right)$
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Apply the power rule for integration, $\displaystyle\int x^n dx=\frac{x^{n+1}}{n+1}$, where $n$ represents a constant function

$\frac{x^{-1}}{-1}-x^{-1}\ln\left(x\right)$
11

Apply the formula: $\frac{x}{-1}$$=-x$, where $x=x^{-1}$

$-x^{-1}\ln\left(x\right)-x^{-1}$
12

Applying the property of exponents, $\displaystyle a^{-n}=\frac{1}{a^n}$, where $n$ is a number

$-\frac{1}{x}\ln\left(x\right)-\frac{1}{x^{1}}$
13

Any expression to the power of $1$ is equal to that same expression

$-\frac{1}{x}\ln\left(x\right)-\frac{1}{x}$
14

$-\frac{1}{x}-\frac{1}{x}\ln\left(x\right)+C_0$

$-\frac{1}{x}-\frac{1}{x}\ln\left(x\right)+C_0$

### Main topic:

Integration by parts

0.28 seconds

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