Indefinite Integration
Integration by Substitution
Grade None
Question:
<p>\(\displaystyle\int\frac{xe^x-1}{x(1+xe^x)}\,dx\) equals</p>
<li>\(\ln\!\left|\dfrac{xe^x}{1+xe^x}\right|+C\)</li>
<li>\(\ln|1+xe^x|-\ln|x|+C\)</li>
<li>\(\ln|x|-\ln|1+xe^x|+C\)</li>
<li>\(\ln|xe^x|-\ln|1+xe^x|+C\)</li>
Step-by-Step Solution
Key Concept: Let u = xeˣ. Then du = (eˣ+xeˣ)dx = eˣ(1+x)dx. The integrand simplifies using this substitution.
<p>Write $\dfrac{xe^x-1}{x(1+xe^x)}$. Multiply top and bottom cleverly:</p>
<p>$$= \frac{d}{dx}\ln|xe^x| - \frac{d}{dx}\ln|1+xe^x|\cdots$$</p>
<p>Check: $\dfrac{d}{dx}\ln|xe^x| = \dfrac{e^x(1+x)}{xe^x}=\dfrac{1+x}{x}=\dfrac{1}{x}+1$.</p>
<p>$\dfrac{d}{dx}\ln|1+xe^x| = \dfrac{e^x(1+x)}{1+xe^x}$.</p>
<p>Sum/difference manipulations give options A, C, D as valid (they differ by constants). Answer: <strong>ACD</strong></p>
Correct Answer: ACD