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  • If <img alt="\mathrm\int \frac{\left(x^{2}+1\right) {e}^{x}}{(x+1)^{2}} {~d} x=f(x) {e}^{x}+C" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%5Cint%20%5Cfrac%7B%5Cleft%28x%5E%7B2%7D

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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