int frac{1+x cos x}{xleft[1-x^2left(e^{sin x} | vedclass

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Question

(D) Let $I = \int \frac{1+x \cos x}{x[1-x^2(e^{\sin x})^2]} dx$. Multiply the numerator and denominator by $e^{\sin x}$: $I = \int \frac{(1+x \cos x)

Vedclass · Accepted Answer

$-\frac{1}{2} \log \left|\frac{\left(x e^{\sin x}\right)^2}{\left(x e^{\sin x}\right)^2-1}\right|+c$

Vedclass · Answer

(D) Let $I = \int \frac{1+x \cos x}{x[1-x^2(e^{\sin x})^2]} dx$. Multiply the numerator and denominator by $e^{\sin x}$: $I = \int \frac{(1+x \cos x) e^{\sin x}}{x e^{\sin x}[1-(x e^{\sin x})^2]} dx$. Let $t = x e^{\sin x}$. Then $dt = [e^{\sin x} + x e^{\sin x} \cos x] dx = e^{\sin x}(1+x \cos x) dx$. Substituting these into the integral: $I = \int \frac{dt}{t(1-t^2)} = \int \frac{dt}{t(1-t)(1+t)}$. Using partial fractions: $\frac{1}{t(1-t)(1+t)} = \frac{1}{t} + \frac{1}{2(1-t)} - \frac{1}{2(1+t)}$. Integrating: $I = \ln|t| - \frac{1}{2} \ln|1-t| - \frac{1}{2} \ln|1+t| + C = \ln|t| - \frac{1}{2} \ln|1-t^2| + C$. $I = \frac{1}{2} \ln|t^2| - \frac{1}{2} \ln|1-t^2| + C = \frac{1}{2} \ln \left| \frac{t^2}{1-t^2} \right| + C$. Substituting back $t = x e^{\sin x}$: $I = \frac{1}{2} \ln \left| \frac{(x e^{\sin x})^2}{1-(x e^{\sin x})^2} \right| + C = -\frac{1}{2} \ln \left| \frac{(x e^{\sin x})^2}{(x e^{\sin x})^2-1} \right| + C$.

$\int \frac{1+x \cos x}{x\left[1-x^2\left(e^{\sin x}\right)^2\right]} d x=$

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