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

Q: Find \displaystyle \int \frac{1}{1+x}\,dx.

\displaystyle \int \frac{1}{1+x}\,dx=\ln|1+x|+C.

Q: What substitution works for \displaystyle \int \frac{1}{1+x}\,dx?

Let u=1+x. Then du=dx, so \displaystyle \int \frac{1}{u}\,du=\ln|u|+C=\ln|1+x|+C.

Q: Compute \displaystyle \int_{0}^{2}\frac{1}{1+x}\,dx.

\displaystyle \left[\ln|1+x|\right]_{0}^{2}=\ln 3-\ln 1=\ln 3.

Q: What is \displaystyle \int \frac{1}{x-1}\,dx?

\displaystyle \int \frac{1}{x-1}\,dx=\ln|x-1|+C since \frac{d}{dx}(x-1)=1.

Q: Check by differentiating \ln|1+x|+C.

\displaystyle \frac{d}{dx}\ln|1+x|=\frac{1}{1+x}. Constant C differentiates to 0.

Answer

To integrate \( \frac{1}{1+x} \), use the fact that \( \int \frac{1}{u}\,du = \ln|u| + C \). Here \(u = 1+x\), so \(du = dx\).

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

Detailed Explanation

We want to find the indefinite integral

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

Step 1: Identify a standard antiderivative form.

Recall the derivative rule for the natural logarithm:

\[
\frac{d}{dx}\left(\ln|u|\right)=\frac{1}{u}\cdot \frac{du}{dx}.
\]

Here, the denominator \(1+x\) suggests choosing \(u=1+x\).

Step 2: Match the integrand with the derivative pattern.

Let

\[
u=1+x.
\]

Differentiate \(u\) with respect to \(x\):

\[
\frac{du}{dx}=1
\quad\Rightarrow\quad
du = dx.
\]

Step 3: Rewrite the integral using substitution.

Because \(du=dx\), the integral becomes

\[
\int \frac{1}{u}\,du.
\]

Step 4: Use the known antiderivative.

We know that

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

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

So the final answer is

\[
\ln|1+x| + C.
\]

Final Answer:

\[
\int \frac{1}{1+x}\,dx = \ln|1+x| + C.
\]

See full solution

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Calculus FAQ

Find \(\displaystyle \int \frac{1}{1+x}\,dx\).

\(\displaystyle \int \frac{1}{1+x}\,dx=\ln|1+x|+C\).

What substitution works for \(\displaystyle \int \frac{1}{1+x}\,dx\)?

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

Why do we use \(\ln|1+x|\) instead of \(\ln(1+x)\)?

The logarithm requires the argument be positive for real values. \(\ln|1+x|\) covers both cases \(1+x>0\) and \(1+x<0\). Derivative still matches \(\frac{1}{1+x}\).

Compute \(\displaystyle \int_{0}^{2}\frac{1}{1+x}\,dx\).

\(\displaystyle \left[\ln|1+x|\right]_{0}^{2}=\ln 3-\ln 1=\ln 3\).

What is \(\displaystyle \int \frac{1}{x-1}\,dx\)?

\(\displaystyle \int \frac{1}{x-1}\,dx=\ln|x-1|+C\) since \(\frac{d}{dx}(x-1)=1\).

Check by differentiating \(\ln|1+x|+C\).

\(\displaystyle \frac{d}{dx}\ln|1+x|=\frac{1}{1+x}\). Constant \(C\) differentiates to \(0\).

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