1. Home
  2. Calculators
  3. Calculators: Calculus II

Integral of ln⁡2(x)\ln^{2}\left(x\right)

The calculator will find the integral/antiderivative of ln⁡2(x)\ln^{2}\left(x\right), with steps shown.

Related calculator: Integral Calculator

Solution

For the integral ∫ln⁡(x)2dx\int{\ln{\left(x \right)}^{2} d x}, use integration by parts ∫u⁡dv⁡=u⁡v⁡−∫v⁡du⁡\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}.

Let u⁡=ln⁡(x)2\operatorname{u}=\ln{\left(x \right)}^{2} and dv⁡=dx\operatorname{dv}=dx.

Then du⁡=(ln⁡(x)2)′dx=2ln⁡(x)xdx\operatorname{du}=\left(\ln{\left(x \right)}^{2}\right)^{\prime }dx=\frac{2 \ln{\left(x \right)}}{x} dx (steps can be seen ») and v⁡=∫1dx=x\operatorname{v}=\int{1 d x}=x (steps can be seen »).

The integral becomes

∫ln⁡(x)2dx=(ln⁡(x)2⋅x−∫x⋅2ln⁡(x)xdx)=(xln⁡(x)2−∫2ln⁡(x)dx){\color{red}{\int{\ln{\left(x \right)}^{2} d x}}}={\color{red}{\left(\ln{\left(x \right)}^{2} \cdot x-\int{x \cdot \frac{2 \ln{\left(x \right)}}{x} d x}\right)}}={\color{red}{\left(x \ln{\left(x \right)}^{2} - \int{2 \ln{\left(x \right)} d x}\right)}}

Apply the constant multiple rule ∫cf(x) dx=c∫f(x) dx\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx with c=2c=2 and f(x)=ln⁡(x)f{\left(x \right)} = \ln{\left(x \right)}:

xln⁡(x)2−∫2ln⁡(x)dx=xln⁡(x)2−(2∫ln⁡(x)dx)x \ln{\left(x \right)}^{2} - {\color{red}{\int{2 \ln{\left(x \right)} d x}}} = x \ln{\left(x \right)}^{2} - {\color{red}{\left(2 \int{\ln{\left(x \right)} d x}\right)}}

For the integral ∫ln⁡(x)dx\int{\ln{\left(x \right)} d x}, use integration by parts ∫u⁡dv⁡=u⁡v⁡−∫v⁡du⁡\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}.

Let u⁡=ln⁡(x)\operatorname{u}=\ln{\left(x \right)} and dv⁡=dx\operatorname{dv}=dx.

Then du⁡=(ln⁡(x))′dx=dxx\operatorname{du}=\left(\ln{\left(x \right)}\right)^{\prime }dx=\frac{dx}{x} (steps can be seen ») and v⁡=∫1dx=x\operatorname{v}=\int{1 d x}=x (steps can be seen »).

The integral becomes

xln⁡(x)2−2∫ln⁡(x)dx=xln⁡(x)2−2(ln⁡(x)⋅x−∫x⋅1xdx)=xln⁡(x)2−2(xln⁡(x)−∫1dx)x \ln{\left(x \right)}^{2} - 2 {\color{red}{\int{\ln{\left(x \right)} d x}}}=x \ln{\left(x \right)}^{2} - 2 {\color{red}{\left(\ln{\left(x \right)} \cdot x-\int{x \cdot \frac{1}{x} d x}\right)}}=x \ln{\left(x \right)}^{2} - 2 {\color{red}{\left(x \ln{\left(x \right)} - \int{1 d x}\right)}}

Apply the constant rule ∫c dx=cx\int c\, dx = c x with c=1c=1:

xln⁡(x)2−2xln⁡(x)+2∫1dx=xln⁡(x)2−2xln⁡(x)+2xx \ln{\left(x \right)}^{2} - 2 x \ln{\left(x \right)} + 2 {\color{red}{\int{1 d x}}} = x \ln{\left(x \right)}^{2} - 2 x \ln{\left(x \right)} + 2 {\color{red}{x}}

Therefore,

∫ln⁡(x)2dx=xln⁡(x)2−2xln⁡(x)+2x\int{\ln{\left(x \right)}^{2} d x} = x \ln{\left(x \right)}^{2} - 2 x \ln{\left(x \right)} + 2 x

Simplify:

∫ln⁡(x)2dx=x(ln⁡(x)2−2ln⁡(x)+2)\int{\ln{\left(x \right)}^{2} d x} = x \left(\ln{\left(x \right)}^{2} - 2 \ln{\left(x \right)} + 2\right)

Add the constant of integration:

∫ln⁡(x)2dx=x(ln⁡(x)2−2ln⁡(x)+2)+C\int{\ln{\left(x \right)}^{2} d x} = x \left(\ln{\left(x \right)}^{2} - 2 \ln{\left(x \right)} + 2\right)+C

Answer

∫ln⁡2(x) dx=x(ln⁡2(x)−2ln⁡(x)+2)+C\int \ln^{2}\left(x\right)\, dx = x \left(\ln^{2}\left(x\right) - 2 \ln\left(x\right) + 2\right) + CA