Indefinite Integration
Integral Calculus-1
star_batch_jee_advanced_2025
Grade 12
Question:
If $\int\frac{(\cos x - \sin x + 1 - x)}{e^x + \sin x + x}dx = \ln(f(x)) + g(x) + c$ where $c$ is the constant of integration & $f(x)$ is positive, then $\frac{f(x) + g(x)}{e^x + \sin x}$ is
Step-by-Step Solution
Key Concept: Decompose the numerator as the derivative of the denominator plus a remainder term to separate the integral into a logarithmic part and a simple part.
Recognize that the integrand $\frac{(e^x+\cos x+1)-(e^x+\sin x+x)}{(e^x+\sin x+x)}$ is of the form $\frac{f(x)+g(x)}{e^x+\sin x+x}$ where $f(x) = e^x+\sin x+x$ and $g(x) = -x$. This can be split as $\frac{f'(x)}{f(x)} + \frac{g(x)}{e^x+\sin x+x}$. The integral becomes $\ln|e^x+\sin x+x| - x + c$ since $\frac{d}{dx}(e^x+\sin x+x) = e^x+\cos x+1$.
Correct Answer: 1