Q. \[ \int \ln\left(x^2\right)\,dx \]

Problem: Find the integral

\[
\int \ln(x^2)\,dx
\]

Step 1: Simplify the logarithm

Recall the logarithm property (valid for positive arguments):

\[
\ln(a^b)=b\ln(a)
\]

Here, \(x^2>0\) for \(x\neq 0\), so for any \(x\neq 0\) we can write:

\[
\ln(x^2)=\ln\bigl((x)^2\bigr)=2\ln|x|
\]

So the integral becomes

\[
\int \ln(x^2)\,dx=\int 2\ln|x|\,dx=2\int \ln|x|\,dx
\]

Step 2: Compute \(\int \ln|x|\,dx\) by integration by parts

Use integration by parts:

\[
\int u\,dv=u\,v-\int v\,du
\]

Choose

\[
u=\ln|x| \quad \text{and} \quad dv=dx
\]

Then compute \(du\) and \(v\):

\[
du=\frac{1}{x}\,dx
\]
\[
v=\int dx=x
\]

Substitute into the integration by parts formula:

\[
\int \ln|x|\,dx=\ln|x|\cdot x-\int x\cdot \frac{1}{x}\,dx
\]

Simplify inside the integral:

\[
\int \ln|x|\,dx=x\ln|x|-\int 1\,dx
\]

Integrate \(1\):

\[
\int 1\,dx=x
\]

So

\[
\int \ln|x|\,dx=x\ln|x|-x+C
\]

Step 3: Multiply by the factor \(2\)

Recall from Step 1 that

\[
\int \ln(x^2)\,dx=2\int \ln|x|\,dx
\]

Substitute the result from Step 2:

\[
\int \ln(x^2)\,dx=2\left(x\ln|x|-x+C\right)
\]

Distribute the \(2\):

\[
\int \ln(x^2)\,dx=2x\ln|x|-2x+2C
\]

Since \(2C\) is just another constant, rename it as \(C\):

\[
\int \ln(x^2)\,dx=2x\ln|x|-2x+C
\]

Final Answer

\[
\int \ln(x^2)\,dx=2x\ln|x|-2x+C
\]