Indefinite Integration
Integration by Parts
Grade None
Question:
<p>[JEE Main 2022] \(\displaystyle\int\frac{\ln x}{(1+x)^2}\,dx\) equals (where \(C\) is constant)</p>
<li>\(-\dfrac{\ln x}{1+x}+\ln|1+x|+C\)</li>
<li>\(\dfrac{\ln x}{1+x}+\ln\!\left|\dfrac{x}{1+x}\right|+C\)</li>
<li>\(-\dfrac{\ln x}{1+x}+\ln\!\left|\dfrac{x}{1+x}\right|+C\)</li>
<li>\(\dfrac{\ln x}{1+x}+\ln|1+x|+C\)</li>
Step-by-Step Solution
Key Concept: Integration by parts: u=lnx, dv=dx/(1+x)^2. Then v=-1/(1+x) and du=dx/x. Integrate the remaining \int1/(x(1+x)) dx by partial fractions.
<p><strong>By parts:</strong> \(u=\ln x,\;dv=\dfrac{dx}{(1+x)^2}\Rightarrow v=-\dfrac{1}{1+x}\).</p>
<p>\[I = -\frac{\ln x}{1+x}+\int\frac{1}{x(1+x)}\,dx\]</p>
<p>\[\int\frac{dx}{x(1+x)} = \int\left(\frac{1}{x}-\frac{1}{1+x}\right)dx = \ln|x|-\ln|1+x|=\ln\!\left|\frac{x}{1+x}\right|\]</p>
<p>\[I = -\frac{\ln x}{1+x}+\ln\!\left|\frac{x}{1+x}\right|+C\]</p>
<p>Answer: <strong>(C)</strong></p>
Correct Answer: C