CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2010

  • question_answer
    \[\int{\frac{\cos x+x\sin x}{{{x}^{2}}+x\cos x}}\]is equal to

    A)  \[\log \left| \frac{\sin x}{1+\cos x} \right|+c\]

    B)  \[\log \left| \frac{\sin x}{x+\cos x} \right|+c\]

    C)  \[\log \left| \frac{2\sin x}{x+\cos x} \right|+c\]

    D)  \[\log \left| \frac{x\sin x}{x+\cos x} \right|+c\]

    E)  \[\log \left| \frac{x}{x+\cos x} \right|+c\]

    Correct Answer: E

    Solution :

    \[\frac{\cos x+x\sin x}{{{x}^{2}}+x\cos x}=\frac{(x+\cos x)-x+x\sin x}{x(x+\cos x)}\] \[=\frac{1}{x}-\frac{(1-\sin x)}{(x+\cos x)}\] \[\therefore \]\[\int{\frac{\cos x+x\sin x}{{{x}^{2}}+x\cos x}}dx\]                                 \[\int{\left( \frac{1}{x}-\frac{1-\sin x}{x+\cos x} \right)}dx\]                 \[=\log x-\log (x+\cos x)+c\]                 \[=\log \left( \frac{x}{x+\cos x} \right)+c\]


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