The value of int {frac{{{e^x} + 9cos x - 2sin | vedclass

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Question

(A) Let $I = \int \frac{e^x + 9\cos x - 2\sin x + 7}{e^x + 7\sin x + 11\cos x + 14} dx$. We express the numerator as $A(e^x + 7\sin x + 11\cos x + 14)

Vedclass · Accepted Answer

$\frac{1}{2}\left( {x + \ln \left( {{e^x} + 7\sin x + 11\cos x + 14} \right)} \right) + C$

Vedclass · Answer

(A) Let $I = \int \frac{e^x + 9\cos x - 2\sin x + 7}{e^x + 7\sin x + 11\cos x + 14} dx$. We express the numerator as $A(e^x + 7\sin x + 11\cos x + 14) + B(e^x + 7\cos x - 11\sin x) + K$. Comparing coefficients: $A + B = 1$ (for $e^x$) $7A - 11B = -2$ (for $\sin x$) $11A + 7B = 9$ (for $\cos x$) $14A = 7 \implies A = \frac{1}{2}$. Substituting $A = \frac{1}{2}$ into $A + B = 1$,we get $B = \frac{1}{2}$. Thus,the numerator is $\frac{1}{2}(e^x + 7\sin x + 11\cos x + 14) + \frac{1}{2}(e^x + 7\cos x - 11\sin x)$. $I = \int \left( \frac{1}{2} + \frac{1}{2} \frac{e^x + 7\cos x - 11\sin x}{e^x + 7\sin x + 11\cos x + 14} \right) dx$. $I = \frac{1}{2}x + \frac{1}{2} \ln |e^x + 7\sin x + 11\cos x + 14| + C$.

The value of $\int {\frac{{{e^x} + 9\cos x - 2\sin x + 7}}{{{e^x} + 7\sin x + 11\cos x + 14}}\,dx} $ is (where $C$ is the constant of integration).

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