∫ex.x3+x+1(1+x2)3/2dx=

I=∫ex.x3+x+1(1+x2)3/2dx=∫exx(x2+1)+1(1+x2)3/2 dx=∫exx(1+x2)1/2+1(1+x2)3/2 dx=∫ex(f(x)+f'(x)) dx where f(x)=x(1+x2)3/2 ⇒f'(x)=(1+x2)1/2−x(1+x2)−1/2(1+x2)1/22 = 1+x2−x21+x2(1+x2)=(1+x2)−x2(1+x2)1+x2=x(1+x2)1/2I=exf(x)+c =exx(1+x2)1/2+c

∫ex.x3+x+1(1+x2)3/2dx=

I=∫ex.x3+x+1(1+x2)3/2dx=∫exx(x2+1)+1(1+x2)3/2 dx=∫exx(1+x2)1/2+1(1+x2)3/2 dx=∫ex(f(x)+f'(x)) dx where f(x)=x(1+x2)3/2 ⇒f'(x)=(1+x2)1/2−x(1+x2)−1/2(1+x2)1/22 = 1+x2−x21+x2(1+x2)=(1+x2)−x2(1+x2)1+x2=x(1+x2)1/2I=exf(x)+c =exx(1+x2)1/2+c

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$\int e^{x} . \frac{x^{3} + x + 1}{{(1 + x^{2})}^{3 / 2}} d x =$

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$e^{x} . \frac{x}{\sqrt{1 + x^{2}}} + c$

$\frac{e^{x}}{\sqrt{1 + x^{2}}} + c$

$\frac{e^{x} \sqrt{1 + x^{2}}}{x} + c$

$\frac{e^{x} . x}{{(1 + x^{2})}^{3 / 2}} + c$

answer is C.

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Detailed Solution

$I = \int e^{x} . \frac{x^{3} + x + 1}{{(1 + x^{2})}^{3 / 2}} d x$

$= \int e^{x} (\frac{x (x^{2} + 1) + 1}{{(1 + x^{2})}^{3 / 2}}) d x$

$= \int e^{x} (\frac{x}{{(1 + x^{2})}^{1 / 2}} + \frac{1}{{(1 + x^{2})}^{3 / 2}}) d x$

$= \int e^{x} (f (x) + f' (x)) d x w h e r e f (x) = \frac{x}{{(1 + x^{2})}^{3 / 2}} \Rightarrow f' (x) = \frac{{(1 + x^{2})}^{1 / 2} - x {(1 + x^{2})}^{- 1 / 2}}{{[{(1 + x^{2})}^{1 / 2}]}^{2}} = \frac{\sqrt{1 + x^{2}} - \frac{x^{2}}{\sqrt{1 + x^{2}}}}{(1 + x^{2})} = \frac{(1 + x^{2}) - x^{2}}{(1 + x^{2}) \sqrt{1 + x^{2}}} = \frac{x}{{(1 + x^{2})}^{1 / 2}}$