Question

$\int\frac{\log(1+x)}{(1+x)}dx=$ ________.

• A. $\frac{1}{2}\log(1+x)+C$
• B. $\frac{1}{2}[\log(1+x)]^{2}+C$
• C. $[\log(1+x)]^{2}+C$
• D. $\log(1+x)+C$
• E. $x \log(1+x)+C$

Hint:

If an integral is of the form $\int f(x) f'(x) dx$, the answer is $\frac{[f(x)]^2}{2} + C$.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
This integral can be solved using the method of substitution. We look for a part of the integrand whose derivative is also present (up to a constant factor). Here, the derivative of \(\log(1+x)\) is \(\frac{1}{1+x}\), which is present in the expression.
Step 2: Key Formula or Approach
1. Let \(u = \log(1+x)\).
2. Find the differential \(du = \frac{d}{dx}(\log(1+x)) \, dx\).
3. Substitute \(u\) and \(du\) into the integral to express it entirely in terms of \(u\).
4. Integrate the resulting simpler expression with respect to \(u\).
5. Substitute back \(u = \log(1+x)\) to get the final answer in terms of \(x\).
Step 3: Detailed Explanation
1. Choose the substitution.
Let \(u = \log(1+x)\).
2. Find du.
Differentiate \(u\) with respect to \(x\):
\[ \frac{du}{dx} = \frac{1}{1+x} \] Rearrange to find the differential \(du\):
\[ du = \frac{1}{1+x} dx \] 3. Substitute into the integral.
The original integral is \(\int \frac{\log(1+x)}{1+x} dx\). We can rewrite this as \(\int \log(1+x) \cdot \frac{1}{1+x} dx\).
Now substitute \(u\) and \(du\):
\[ \int u \, du \] 4. Integrate with respect to u.
Using the power rule for integration (\(\int u^n \, du = \frac{u^{n+1}}{n+1}\)):
\[ \int u^1 \, du = \frac{u^{1+1}}{1+1} + C = \frac{u^2}{2} + C \] 5. Substitute back for x.
Replace \(u\) with \(\log(1+x)\):
\[ \frac{(\log(1+x))^2}{2} + C = \frac{1}{2}[\log(1+x)]^2 + C \] Step 4: Final Answer
The integral is \(\frac{1}{2}[\log(1+x)]^2 + C\).