What is the integral of ln^2(x)?

integral of ln^2(x)

The integral of

\ln^2(x)

x\ln^2(x) - 2x\ln(x) + 2x + C

.

Solution. We want to determine the integral of

\ln^2(x)

, i.e.:

\begin{align*} \int \ln^2(x) dx. \end{align*}

We need to integrate by parts; that is, we will use the following formula:

\begin{align*} \int U dV = UV - \int VdU, \end{align*}

where

U = \ln^2(x)

and

dU = 1 \cdot dx

. Then we get

dU = \frac{2\ln(x)}{x}

which we saw here, and

V = x

. So we get:

\begin{align*} \int \ln^2(x) dx &= \ln^2(x)\cdot x - \int x \cdot \frac{2\ln(x)}{x}dx \\ &= x\ln^2(x) - \int 2\ln(x) dx. \end{align*}

Now we want to integrate

2\ln(x)

into parts, i.e.:

\int 2\ln(x) dx

. Let

U' = \ln(x)

and

dV' = 2dx

. Then we saw here that

dU' = \frac{1}{x}

, and

V' = 2x

. So we get:

\begin{align*} \int \ln^2(x) dx &= x\ln^2(x) - \int 2\ln(x) dx \\ &= x\ln^2(x) - (2x\ln(x) - \int \frac{2x}{x} dx )\\ &= x\ln^2(x) - 2x\ln(x) + \int 2 dx \\ &= x\ln^2(x) - 2x\ln(x) + 2x + C. \end{align*}

Therefore, the integral of

\ln^2(x)

x\ln^2(x) - 2x\ln(x) + 2x + C