Let - ex f x -f- x dx- x - Then - exf x dx is

The correct option is B 1/2{ϕ ( x )+e^xf ( x ) } ∫ e ^ x f( x ) dx= e ^ x f( x ) ∫ e ^ x f'( x ) dx = e ^ x f( x ) +ϕ( x ) ∫ e ^ x f( x ) ...

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Integration Using Substitution

Let ∫ ex f ...

Question

Let $\int e^{x} {f (x) - f^{'} (x)} d x = ϕ (x) .$ Then $\int e^{x} f (x) d x$ is

ϕ (x) + e^{x} f (x)

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ϕ (x) - e^{x} f (x)

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\frac{1}{2} {ϕ (x) + e^{x} f (x)}

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\frac{1}{2} {ϕ (x) + e^{x} f^{'} (x)}

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Solution

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Similar questions

\int e^{x} {f (x) - f^{'} (x)} d x = ϕ (x) .

\int e^{x} f (x) d x

\int e^{x} (f (x) - f^{'} (x)) d x = ϕ (x)

\int e^{x} f (x) d x =

\int e^{x} {f (x) - f^{'} (x)} d x = ϕ (x)

\int e^{x} f (x) d x

c

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

\int e^{x} (f (x) + f^{'} (x)) d x = e^{x} f (x) + C

\int e^{x} (tan x + {sec}^{2} x) d x = e^{x} tan x + C

\int e^{x} (f (x) + f^{'} (x)) d x = e^{x} f (x) + C

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