Indefinite IntegrationhardFree

Find f(x) from ∫(x+1)/(x(1+xeˣ)²)dx | JEE

JEE Maths question with a full step-by-step solution.

Question
If ∫x+1x(1+xex)2 dx=log⁡∣1−f(x)∣+f(x)+C\displaystyle\int \dfrac{x+1}{x(1+xe^{x})^{2}}\,dx=\log|1-f(x)|+f(x)+C, then f(x)f(x) equals
A1x+ex\dfrac{1}{x+e^{x}}
B11+xex\dfrac{1}{1+xe^{x}}correct
C1(1+xex)2\dfrac{1}{(1+xe^{x})^{2}}
D1(1+ex)2\dfrac{1}{(1+e^{x})^{2}}
Solution
Step 1: The factor (1+xex)(1+xe^{x}) suggests the substitution t=xext=xe^{x}. Then
dt=(ex+xex) dx=(x+1)ex dxdt=(e^{x}+xe^{x})\,dx=(x+1)e^{x}\,dx
 Step 2: Introduce exe^{x} to match dtdt. Multiply numerator and denominator by exe^{x}:
I=∫(x+1)exxex(1+xex)2 dx=∫dtt(1+t)2I=\int\dfrac{(x+1)e^{x}}{xe^{x}(1+xe^{x})^{2}}\,dx=\int\dfrac{dt}{t(1+t)^{2}}
 Step 3: Resolve into partial fractions. Writing 1t(1+t)2=At+B1+t+D(1+t)2\dfrac{1}{t(1+t)^{2}}=\dfrac{A}{t}+\dfrac{B}{1+t}+\dfrac{D}{(1+t)^{2}} gives A=1, B=−1, D=−1A=1,\ B=-1,\ D=-1:
1t(1+t)2=1t−11+t−1(1+t)2\dfrac{1}{t(1+t)^{2}}=\dfrac{1}{t}-\dfrac{1}{1+t}-\dfrac{1}{(1+t)^{2}}
 Step 4: Integrate each term:
I=ln⁡∣t∣−ln⁡∣1+t∣+11+t+C=ln⁡∣t1+t∣+11+t+CI=\ln|t|-\ln|1+t|+\dfrac{1}{1+t}+C=\ln\left|\dfrac{t}{1+t}\right|+\dfrac{1}{1+t}+C
 Step 5: Match the required form. Since t1+t=1−11+t\dfrac{t}{1+t}=1-\dfrac{1}{1+t},
I=log⁡∣1−11+t∣+11+t+CI=\log\left|1-\dfrac{1}{1+t}\right|+\dfrac{1}{1+t}+C
 Comparing with log⁡∣1−f(x)∣+f(x)+C\log|1-f(x)|+f(x)+C gives f(x)=11+t=11+xexf(x)=\dfrac{1}{1+t}=\dfrac{1}{1+xe^{x}}. Correct answer: (2)
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