$\int \frac{\left(x^{2}+1\right) e^{x}}{(x+1)^{2}} d x=f(x) e^{x}+C$, Where $C$ is a constant, then $\frac{d^{3} f}{d x^{3}}$ at $x=1$ is equal to :

JEE Main · 2022 · 27 Jun 2022 · Shift 1 · Indefinite Integrals

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

, Where

C

is a constant, then

\frac{d ^{3} f}{d x ^{3}}

x = 1

is equal to :

Answer: (B)

Rewrite the integrand:

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

Let

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

. Then

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

.

The integral becomes:

\int (f (x) + f^{'} (x)) e^{x} d x = f (x) e^{x} + C

Now compute derivatives of

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

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

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

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

x = 1

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

JEE Main · 2022 · 27 Jun 2022 · Shift 1 · Single correct option