If \int (x^2 + 1)e^x \over (x + 1)^2dx = f(x)e^x + C , where C is a constant, then d^3f \over dx^3 at x = 1 is equal to :

Indefinite Integrals PYQ — JEE Main 2022 · 27 Jun · Shift 1 · Q26 with Solution

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Correct answer: B

We are given

\int \frac{(x^2+1)e^x}{(x+1)^2}\,dx = f(x)e^x + C.

We need to find

f'''(1).

Differentiate the right-hand side:

\frac{d}{dx}\big(f(x)e^x\big)=e^x\big(f(x)+f'(x)\big).

Since this must equal the integrand,

e^x\big(f+f'\big)=\frac{(x^2+1)e^x}{(x+1)^2}.

Cancel $e^x$ :

f+f' = \frac{x^2+1}{(x+1)^2}.

So we must solve

f'+f = \frac{x^2+1}{(x+1)^2}.

Notice that

\frac{x^2+1}{(x+1)^2} = 1 - \frac{2x}{(x+1)^2}.

But a better approach is to guess a simple rational function for $f(x)$ .

Try

f(x)=\frac{x-1}{x+1}.

Then

f'(x)=\frac{(x+1)-(x-1)}{(x+1)^2}=\frac{2}{(x+1)^2}.

Hence

f+f' = \frac{x-1}{x+1}+\frac{2}{(x+1)^2} =\frac{(x-1)(x+1)+2}{(x+1)^2} =\frac{x^2-1+2}{(x+1)^2} =\frac{x^2+1}{(x+1)^2}.

So this works.

Thus we can take

f(x)=\frac{x-1}{x+1}.

Now compute derivatives.

First derivative:

f'(x)=\frac{2}{(x+1)^2}=2(x+1)^{-2}.

Second derivative:

f''(x)=2(-2)(x+1)^{-3}=-\frac{4}{(x+1)^3}.

Third derivative:

f'''(x)=-4(-3)(x+1)^{-4}=\frac{12}{(x+1)^4}.

Evaluate at $x=1$ :

f'''(1)=\frac{12}{(2)^4}=\frac{12}{16}=\frac{3}{4}.

Therefore the correct option is

\boxed{\frac{3}{4}}

which is option B.

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