- ex x2 - 1 x-1 2dx is equal to

I= ∫ e^x (x^2 + 1 )(x+1 )^2dx =∫e^x (x^2 1+2 )(x+1 )^2dx =∫ e^x [ x 1x+1 + 2(x+1 )^2 ]dx =∫ e^x[f(x )+f'(x ) ]dx, where f(x )=x 1x+1 and f'(x )=2( ...

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Byju's Answer

Standard VIII

Mathematics

Division of a Polynomial by a Polynomial

∫ ex x2 + 1 x...

Question

$\int e^{x} \frac{(x^{2} + 1)}{{(x + 1)}^{2}} d x$ is equal to

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

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

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e^{x} (x + 1) (x - 1) + C

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\frac{e^{x}}{(x + 1)} + C

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Solution

The correct option is A $\frac{(x - 1)}{(x + 1)} e^{x} + C$
$I = \int e^{x} \frac{(x^{2} + 1)}{{(x + 1)}^{2}} d x = \int \frac{e^{x} (x^{2} - 1 + 2)}{{(x + 1)}^{2}} d x = \int e^{x} [\frac{x - 1}{x + 1} + \frac{2}{{(x + 1)}^{2}}] d x = \int e^{x} [f (x) + f^{'} (x)] d x, where f (x) = \frac{x - 1}{x + 1} and f^{'} (x) = \frac{2}{{(x + 1)}^{2}} {∵ \int e^{x} {f (x) + f^{'} (x)} d x = e^{x} f (x) + C} \Rightarrow I = e^{x} (\frac{x - 1}{x + 1}) + C$

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Similar questions

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

is equal to

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

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

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

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

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Division of a Polynomial by a Polynomial

Standard VIII Mathematics

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∫ex(x2+1)(x+1)2dx is equal to

The correct option is A (x−1)(x+1)ex+CI=∫ex(x2+1)(x+1)2dx=∫ex(x2−1+2)(x+1)2dx=∫ex[x−1x+1+2(x+1)2]dx=∫ex[f(x)+f′(x)]dx,where f(x)=x−1x+1 and f′(x)=2(x+1)2{∵∫ex{f(x)+f′(x)}dx=exf(x)+C}⇒I=ex(x−1x+1)+C

$\int e^{x} \frac{(x^{2} + 1)}{{(x + 1)}^{2}} d x$ is equal to