Indefinite IntegrationmediumFree

Indefinite Integration: 2dx Constant Integration

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

Question

I = \displaystyle\int\left(x^2+1\right)\left(\left(x+1\right)e^x\right)^2dx = A\left(f(x)\right)^2+c

, where

c

is the constant of integration and

f(-1) = \dfrac2e

, then

2A+f(0)

Solution

Answer: 2

Step 1:

\left(x^2+1\right)\left(\left(x+1\right)e^x\right)^2 = \left(x^2+1\right)\left(x+1\right)^2e^{2x} = \underbrace{\left(x^2+1\right)e^x}_{\text{call this } t}\cdot\underbrace{\left(x+1\right)^2e^x}_{\text{this is } dt/dx}

t = \left(x^2+1\right)e^x \quad\Longrightarrow\quad \frac{dt}{dx} = 2xe^x+\left(x^2+1\right)e^x = \left(x^2+2x+1\right)e^x = \left(x+1\right)^2e^x .

Step 2:

I = \int t\,dt = \frac{t^2}2+c = \frac12\left[\left(x^2+1\right)e^x\right]^2+c

A = \frac12 ,\qquad f(x) = \left(x^2+1\right)e^x

Step 3:

f(-1) = \left(1+1\right)e^{-1} = \frac2e .

Both

f

and

-f

give the same

\left(f(x)\right)^2

; this value picks the

+

branch, so

f

is fixed. Step 4:

f(0) = \left(0+1\right)e^0 = 1 ,\qquad 2A+f(0) = 2\cdot\frac12+1 = 2

Answer:

2

Still stuck on this question?Ask your doubt on WhatsApp

Solve more, learn faster

Sign up free to solve more JEE Maths questions and explore doMath — timed drills, mastery sprints, bookmarks, and chapter-wise progress tracking.