Question Statement
Q11. Evaluate the integral:
∫(ex+1)2(ex−2)exdx
Background and Explanation
This problem integrates a function involving exponential expressions in the denominator. The strategy is to first use substitution to convert the integrand into a proper rational function, then apply partial fraction decomposition to split it into simpler integrals.
Solution
Let
I=∫(ex+1)2(ex−2)exdx
Put ex=t, which gives exdx=dt.
The integral becomes:
I=∫(t+1)2(t−2)dt
Decompose the integrand:
(t+1)2(t−2)1=t+1A+(t+1)2B+t−2C
Multiplying both sides by (t+1)2(t−2):
1=A(t+1)(t−2)+B(t−2)+C(t+1)2
Put t=−1 (i.e., t+1=0) in equation (3):
1=A(0)(−3)+B(−3)+C(0)2
1=−3B⟹B=3−1
Put t=2 (i.e., t−2=0) in equation (3):
1=A(3)(0)+B(0)+C(3)2
1=9C⟹C=91
Compare coefficients of t2 by expanding equation (3):
1=(A+C)t2+(−A+B+2C)t+(−2A−2B+C)
Coefficient of t2: 0=A+C=A+91
⟹A=9−1
Substituting A, B, C into (2) and integrating:
I=9−1∫t+1dt−31∫(t+1)−2dt+91∫t−2dt
=9−1ln∣t+1∣−31⋅−1(t+1)−1+91ln∣t−2∣+C
=9−1ln(t+1)+3(t+1)1+91ln(t−2)+C
Substituting t=ex:
I=9−1ln(ex+1)+3(ex+1)1+91ln(ex−2)+C
| Formula | Application |
|---|
| ex=t⇒exdx=dt | Substitution to rationalize |
| (t+1)2(t−2)1=t+1A+(t+1)2B+t−2C | Repeated linear factor decomposition |
| $\int \frac,dx = \ln | x |
| ∫xndx=n+1xn+1+C | Power rule (used for (t+1)−2) |
Summary of Steps
- Substitute ex=t to get ∫(t+1)2(t−2)dt
- Set up partial fractions: t+1A+(t+1)2B+t−2C
- Find constants: B=−31 (put t=−1), C=91 (put t=2), A=−91 (compare t2 coefficients)
- Integrate each term separately
- Substitute back t=ex to obtain the final answer