JEE MATHEMATICS MCQ: INDEFINITE INTEGRATION

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\int\left(\frac{e^{\sin x}}{e^{\sin x}+x}\right)^2\left(\frac{1}{x}-\cos x\right) d x=\log _e(f(x))+g(x)+c

; where

c

is constant, then

f(\pi)-g(\pi)

View answer and explanation

Correct option: B

\frac{\pi-1}{\pi+1}

Option B is correct because after the substitution and integration, the expression simplifies to the given form, and evaluating f(π) - g(π) yields (π-1)/(π+1).

More Questions from INDEFINITE INTEGRATION

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\int \sin ^5 x \cos ^4 x d x=

f(x)

is differentibale and

\int\limits_{0}^{t^2} x f(x) d x=\frac{2}{5} t^5

then

f\left(\frac{4}{25}\right)

equals If

\int \frac{1-7 \cos ^2 x}{\sin ^7 x \cos ^2 x} d x=\frac{f(x)}{\sin ^7 x}+C

is equal to If

\int {e}^{{x}^2}\left(2-\frac{1}{{x}^2}\right) {de}={e}^{{x}^2} {f}({x})+{C}

and

{f}\left(\frac{1}{2}\right)=2

, then

f(1)

is equal to (where

{C}

is an arbitrary constant)

\int \frac{x^7}{\left(1-x^2\right)^5} d x

is equal to:

\int \,\frac{{a{x^3} + \,\,b{x^2} + c}}{{{x^4}}}\;\,dx

equals