Q. \[ \int \frac{x}{x^2+1}\,dx \]

Q: How do I integrate by substitution for \int \frac{x}{x^2+1}\,dx?

Let u=x^2+1. Then du=2x\,dx, so \int \frac{x}{x^2+1}\,dx=\frac12\int \frac{1}{u}\,du=\frac12\ln|u|+C=\frac12\ln(x^2+1)+C.

Q: Can I verify the result by differentiating \frac12\ln(x^2+1)?

Differentiate: \frac{d}{dx}\left(\frac12\ln(x^2+1)\right)=\frac12\cdot\frac{2x}{x^2+1}=\frac{x}{x^2+1}.

Q: What u-substitution should I choose when the denominator is x^2+1?

Choose u=x^2+1 because du=2x\,dx matches the numerator x\,dx.

Q: Why does the logarithm appear in \int \frac{x}{x^2+1}\,dx?

Because after substitution the integrand becomes \int \frac{1}{u}\,du, and \int \frac{1}{u}\,du=\ln|u|+C.

Q: Do I need absolute value in the final answer?

Yes: \ln|x^2+1|. Since x^2+1>0 for all real x, you can also write \ln(x^2+1) without changing the function.

Q: How do I solve it using partial fractions or other methods?

Partial fractions are unnecessary. Recognize it as \int \frac{1}{2}\cdot\frac{2x}{x^2+1}\,dx, then use u=x^2+1 to get \frac12\ln(x^2+1)+C.

Answer

We compute

\[
\int \frac{x}{x^2+1}\,dx
\]

Let \(u=x^2+1\). Then \(du=2x\,dx\), so \(x\,dx=\frac{1}{2}du\).

\[
\int \frac{x}{x^2+1}\,dx=\frac{1}{2}\int \frac{1}{u}\,du=\frac{1}{2}\ln|u|+C
\]

Substitute back \(u=x^2+1\).

\[
\frac{1}{2}\ln\left|x^2+1\right|+C
\]

Detailed Explanation

We want to compute the integral

\[
\int \frac{x}{x^2+1}\,dx .
\]

Step 1: Choose a substitution.

Look at the denominator \(x^2+1\). Its derivative is related to the numerator:

\[
\frac{d}{dx}\left(x^2+1\right)=2x.
\]

Since the numerator is \(x\), this suggests using the substitution

\[
u=x^2+1.
\]

Step 2: Compute \(du\).

Differentiate both sides:

\[
du = d(x^2+1)=2x\,dx.
\]

This means we can rewrite \(x\,dx\) in terms of \(du\):

\[
x\,dx=\frac{1}{2}\,du.
\]

Step 3: Rewrite the integral in terms of \(u\).

Substitute \(u=x^2+1\) and \(x\,dx=\frac{1}{2}du\):

\[
\int \frac{x}{x^2+1}\,dx
=
\int \frac{1}{u}\cdot \frac{1}{2}\,du
=
\frac{1}{2}\int \frac{1}{u}\,du.
\]

Step 4: Integrate.

Recall that

\[
\int \frac{1}{u}\,du = \ln|u|+C.
\]

So we get

\[
\frac{1}{2}\int \frac{1}{u}\,du
=
\frac{1}{2}\ln|u|+C.
\]

Step 5: Substitute back \(u=x^2+1\).

\[
\frac{1}{2}\ln|u|+C
=
\frac{1}{2}\ln|x^2+1|+C.
\]

Since \(x^2+1\) is always positive, we can remove the absolute value if desired:

\[
\boxed{\int \frac{x}{x^2+1}\,dx=\frac{1}{2}\ln\left(x^2+1\right)+C.}
\]

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Calculus FAQ

How do I integrate by substitution for \(\int \frac{x}{x^2+1}\,dx\)?

Let \(u=x^2+1\). Then \(du=2x\,dx\), so \(\int \frac{x}{x^2+1}\,dx=\frac12\int \frac{1}{u}\,du=\frac12\ln|u|+C=\frac12\ln(x^2+1)+C\).

Can I verify the result by differentiating \(\frac12\ln(x^2+1)\)?

Differentiate: \(\frac{d}{dx}\left(\frac12\ln(x^2+1)\right)=\frac12\cdot\frac{2x}{x^2+1}=\frac{x}{x^2+1}\).

What \(u\)-substitution should I choose when the denominator is \(x^2+1\)?

Choose \(u=x^2+1\) because \(du=2x\,dx\) matches the numerator \(x\,dx\).

Why does the logarithm appear in \(\int \frac{x}{x^2+1}\,dx\)?

Because after substitution the integrand becomes \(\int \frac{1}{u}\,du\), and \(\int \frac{1}{u}\,du=\ln|u|+C\).

Do I need absolute value in the final answer?

Yes: \(\ln|x^2+1|\). Since \(x^2+1>0\) for all real \(x\), you can also write \(\ln(x^2+1)\) without changing the function.

How do I solve it using partial fractions or other methods?

Partial fractions are unnecessary. Recognize it as \(\int \frac{1}{2}\cdot\frac{2x}{x^2+1}\,dx\), then use \(u=x^2+1\) to get \(\frac12\ln(x^2+1)+C\).

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