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

Option: 1
$-\frac{3}{4}$

Option: 2
$\frac{3}{4}$

Option: 3
$-\frac{3}{2}$

Option: 4
$\frac{3}{2}$

Answers (1)

best_answer

We have to convert this integral into the form $\mathrm{\int e^{x}\left(f(x)+f^{\prime}(x)\right) d x}$ whose

integral is $\mathrm{e^{x} f(1)+c}$ .

So let us assume $\mathrm{f(x)}$ to be

$\mathrm{f(x)=\frac{a x+b}{x+1} \Rightarrow f^{\prime}(x) =\frac{a(x+1)-(a x+b)}{(x+1)^{2}}} \\$

$\mathrm{=\frac{a-b}{(x+1)^{2}}}$

$\mathrm{f(x)+f^{\prime}(x)=\frac{(a x+b)(x+1)+a-b}{(x+1)^{2}}=x^{2}+1} \\$

$\mathrm{\Rightarrow a=1, a+b=0 \Rightarrow b=-1}$

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

$\mathrm{f^{\prime}(x)=\frac{2}{(x+1)^{2}}, f^{\prime \prime}(x)=\frac{-4}{(x+1)^{3}}, f^{\prime \prime \prime}(x): \frac{12}{(x+1)^{4}}}$

$\mathrm{f^{\prime \prime \prime}(1): \frac{12}{(2)^{4}}: \frac{12}{16}=\frac{3}{4}}$

Hence the correct answer is option 2

Posted by

Divya Prakash Singh

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