# What is the antiderivative of ln(x)^2?

Jun 14, 2018

$I = 2 x \left[\ln x - 1\right] + c$

#### Explanation:

Here,

$I = \int \ln {\left(x\right)}^{2} \mathrm{dx}$

Using Power-coefficient Rule :

color(red)(lnx^n=n*lnx , we get

$I = \int \ln {x}^{2} \mathrm{dx} = \int 2 \cdot \ln x \ldots \to \left(A\right)$

Using , Integration by parts:

color(blue)(int(u*v)dx=uintvdx-int(u'intvdx)dx

$u = \ln x \mathmr{and} v = 2 \implies u ' = \frac{1}{x} \mathmr{and} \int v \mathrm{dx} = 2 x$

$I = \ln x \cdot 2 x - \int \frac{1}{x} \cdot 2 x \mathrm{dx} + c$

$\implies I = 2 x \cdot \ln x - \int 2 \mathrm{dx} + c$

$\implies I = 2 x \ln x - 2 x + c$

$\implies I = 2 x \left[\ln x - 1\right] + c$
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Note:

If $I = {\left(\ln x\right)}^{2}$ ,then

$I = {\left(\ln x\right)}^{2} \cdot 1 \mathrm{dx}$

Using , Integration by parts:

$u = {\left(\ln x\right)}^{2} \mathmr{and} v = 1 \implies u ' = \frac{2 \ln x}{x} \mathmr{and} \int v \mathrm{dx} = x$

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

$= {\left(\ln x\right)}^{2} \cdot x - \int 2 \cdot \ln x$

Using above result for $\left(A\right)$

$I = x \cdot {\left(\ln x\right)}^{2} - \left\{2 x \left[\ln x - 1\right]\right\} + c$

Jun 14, 2018

$2 x \ln x - x + C$

#### Explanation:

The antiderivative of a function is basically the function's integral. So we get:

$\int \ln {\left(x\right)}^{2} \setminus \mathrm{dx}$

I'm assuming that we have $\int \ln {x}^{2} \setminus \mathrm{dx}$.

Using logarithm rules, we get:

$= \int 2 \ln x \setminus \mathrm{dx}$

$= 2 \int \ln x \setminus \mathrm{dx}$

This is a common integral, where $\int \ln x \setminus \mathrm{dx} = x \ln x - x + C$.

$\therefore = 2 x \ln x - 2 x + C$