AP EAMCET · Maths · Indefinite Integration
\(\int \frac{1+x \cos x}{x\left[1-x^2\left(e^{\sin x}\right)^2\right]} d x=\)
- A \(\frac{1}{2} \log \left|\frac{\left(x e^{\sin x}\right)^2}{\left(x e^{\sin x}\right)^2+1}\right|+c\)
- B \(-\frac{1}{2} \log \left|\frac{\left(x e^{\sin x}\right)^2}{\left(x e^{\sin x}\right)^2+1}\right|+c\)
- C \(\frac{1}{2} \log \left|\frac{\left(x e^{\sin x}\right)^2}{\left(x e^{\sin x}\right)^2-1}\right|+c\)
- D \(-\frac{1}{2} \log \left|\frac{\left(x e^{\sin x}\right)^2}{\left(x e^{\sin x}\right)^2-1}\right|+c\)
Answer & Solution
Correct Answer
(C) \(\frac{1}{2} \log \left|\frac{\left(x e^{\sin x}\right)^2}{\left(x e^{\sin x}\right)^2-1}\right|+c\)
Step-by-step Solution
Detailed explanation
\[ \begin{aligned} & \text { Let } \mathrm{I}=\int \frac{(1+x \cos x)}{x\left[1-x^2\left(e^{\sin x}\right)^2\right]} \\ & d x=\int \frac{(1+x \cos x) e^{\sin x}}{x e^{\sin x}\left[1-\left(x e^{\sin x}\right)^2\right]} d x \end{aligned} \] Let…
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