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Let $f (x) = \frac{x e^{x}}{(1 + x)^{2}}, x \neq - 1$
Statement-1: The antiderivative $F$ of $f (x)$ satisfying $F (0) = 1$
is $\frac{1}{1 + x} e^{x}$
Statement-2: $f (x)$ is of the form $(g (x) + g^{'} (x)) e^{x}$ for some
$g (x)$ .

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STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is a correct explanation for STATEMENT-1

STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is NOT a correct explanation forSTATEMENT-1

STATEMENT-1 is True, STATEMENT-2 is False

STATEMENT-1 is False, STATEMENT-2 is True

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detailed solution

Correct option is A

F(x)=∫xex(1+x)2dx=∫11+x−1(1+x)2exdx=11+xex+CF(0)=1⇒C=0 so F(x)=11+xex.

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Let f(x)=xex(1+x)2,x≠−1 Statement-1: The antiderivative F of f (x) satisfying F(0)=1is 11+xexStatement-2: f (x) is of the form g(x)+g′(x)ex for some g(x).

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Let $f (x) = \frac{x e^{x}}{(1 + x)^{2}}, x \neq - 1$
Statement-1: The antiderivative $F$ of $f (x)$ satisfying $F (0) = 1$
is $\frac{1}{1 + x} e^{x}$
Statement-2: $f (x)$ is of the form $(g (x) + g^{'} (x)) e^{x}$ for some
$g (x)$ .