Question

∫\(\frac {e^x}{(2+e^x)(e^x +1)}\)dx = (where C is a constant of integration.)

• A. log \((\frac {e^x + 2}{e^x +1})\) + C
• B. log \((\frac {e^x}{e^x +2})\) + C
• C. \((\frac {e^x + 1}{e^x +2})\) + C
• D. log \((\frac {e^x + 1}{e^x +2})\) + C

Answer 1

The Correct Option is D

Detailed Solution:

Step 1: Substitution
Let t = eˣ ⇒ dt = eˣ dx

Given integral becomes:
I = ∫ [t / ((2 + t)(t + 1))] × (1 / t) dt

This simplifies to:
I = ∫ [1 / ((2 + t)(1 + t))] dt

Step 2: Partial Fraction Decomposition
Break the expression into partial fractions:

1 / ((t + 1)(t + 2)) = A / (t + 1) + B / (t + 2)

Multiplying both sides by (t + 1)(t + 2):
1 = A(t + 2) + B(t + 1)

Now solve for A and B:

• Put t = -2 ⇒ 1 = A(0) + B(-1) ⇒ B = -1
• Put t = -1 ⇒ 1 = A(1) + B(0) ⇒ A = 1

So, the integrand becomes:
I = ∫ [1 / (t + 1) - 1 / (t + 2)] dt

Step 3: Integrate

I = ∫ [1 / (t + 1)] dt - ∫ [1 / (t + 2)] dt
I = log|t + 1| - log|t + 2| + C

Using the logarithmic identity:
log A - log B = log (A / B)

I = log((t + 1) / (t + 2)) + C

Step 4: Substitute back t = eˣ

I = log((eˣ + 1) / (eˣ + 2)) + C

Final Answer:
log((eˣ + 1) / (eˣ + 2)) + C

Correct Option: (D)