The correct option is B 1/x+sinx+c x+sin x=t (1+cos x)dx=dt ∫dtt^2= 1t+c = 1(x+sin x)+c ∫1+cos x(x+sin x)^2dx= 1(x+sin x)+c ...

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Special Integrals - 1

∫1+cosxx+sinx...

Question

$\int \frac{1 + c o s x}{(x + s i n x)^{2}} d x =$

\frac{1}{x + s i n x} + c

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\frac{- 1}{x + s i n x} + c

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\frac{1}{(x + s i n x)^{2}} + c

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log | x +

sin

x | + c

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Solution

The correct option is B $\frac{- 1}{x + s i n x} + c$
$x + s i n x = t$
$(1 + c o s x) d x = d t$
$\int \frac{d t}{t^{2}} = \frac{- 1}{t} + c$
$= - \frac{1}{(x + s i n x)} + c$
$\int \frac{1 + c o s x}{(x + s i n x)^{2}} d x = - \frac{1}{(x + s i n x)} + c$

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