The correct option is B e^x/1+x^2+c ∫ e^x(1 x)^2(1+x^2)^2dx ∫ e^x [ (1+x^2) 2x(1+x^2)^2 ]dx ∫ e^x [ 11+x^2 2x(1+x^2)^2 ]dx It is in the for ...

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Proof by mathematical induction

∫ ex1-x/1+x22...

Question

$\int e^{x} (\frac{1 - x}{1 + x^{2}})^{2}$ dx $=$

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

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

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\frac{e^{x}}{(1 + x^{2})^{2}} + c

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

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Solution

The correct option is B $\frac{e^{x}}{1 + x^{2}} + c$
$\int e^{x} \frac{(1 - x)^{2}}{(1 + x^{2})^{2}} d x$
$\int e^{x} [\frac{(1 + x^{2}) - 2 x}{(1 + x^{2})^{2}}] d x$
$\int e^{x} [\frac{1}{1 + x^{2}} - \frac{2 x}{(1 + x^{2})^{2}}] d x$
It is in the form of
$\int e^{x} (f (x) + f^{'} (x)) d x$
$= e^{x} f (x) + c$
$\Rightarrow \int e^{x} [\frac{1}{1 + x^{2}} - \frac{2 x}{1 + x^{2}}] d x = e^{x} \frac{1}{1 + x^{2}} + c$

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\int e^{x} {(\frac{1 - x}{1 + x^{2}})}^{2} d x is equal to (a) \frac{e^{x}}{1 + x} + C (b) - \frac{e^{x}}{1 + x^{2}} + C (c) \frac{e^{x}}{(1 + x}

$\int e^{x} {(\frac{1 - x}{1 + x^{2}})}^{2}$ dx

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

\int \frac{d x}{\sqrt{1 - (x^{2})^{2}}} = {sin}^{- 1} x^{2} + C

\int \frac{d x}{\sqrt{1 - x^{2}}} = {sin}^{- 1} x + C

$\int e^{(x + \frac{1}{x})} (1 + x - \frac{1}{x}) d x = e^{(x + \frac{1}{x})} f (x) + c t h e n \frac{d f (x)}{d x}$

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