IGDTUW 2025 — Mathematics PYQ
IGDTUW | Mathematics | 2025Integrate: ∫((1+x)2xex)dx
Choose the correct answer:
- A.
((1+x)2ex)+C
- B.
(1+xex)+C
(1+xex)+C
Explanation
Solution
To solve this integration, we use the standard calculus identity:
∫ex[f(x)+f′(x)]dx=exf(x)+C
Step 1: Rearrange the Numerator
We can rewrite the x in the numerator as (1+x)−1:
∫(1+x)2((1+x)−1)exdx
Step 2: Split the Fraction
Distribute ex and split the terms over the common denominator:
∫ex[(1+x)21+x−(1+x)21]dx
∫ex[1+x1−(1+x)21]dx
Step 3: Identify f(x) and f′(x)
Let:
-
f(x)=1+x1
-
f′(x)=−(1+x)21
Since the expression is now in the form ∫ex[f(x)+f′(x)]dx, the result is exf(x)+C.
∫ex[1+x1+(−(1+x)21)]dx=1+xex+C
The correct option is (b) (1+xex)+C.
Explanation
Solution
To solve this integration, we use the standard calculus identity:
∫ex[f(x)+f′(x)]dx=exf(x)+C
Step 1: Rearrange the Numerator
We can rewrite the x in the numerator as (1+x)−1:
∫(1+x)2((1+x)−1)exdx
Step 2: Split the Fraction
Distribute ex and split the terms over the common denominator:
∫ex[(1+x)21+x−(1+x)21]dx
∫ex[1+x1−(1+x)21]dx
Step 3: Identify f(x) and f′(x)
Let:
-
f(x)=1+x1
-
f′(x)=−(1+x)21
Since the expression is now in the form ∫ex[f(x)+f′(x)]dx, the result is exf(x)+C.
∫ex[1+x1+(−(1+x)21)]dx=1+xex+C
The correct option is (b) (1+xex)+C.

