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\[ \large \int \dfrac x{1+ x \tan x} \, dx = \, ? \]

Note by Aritra Jana 1 year, 8 months ago

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\[\begin{eqnarray}&& \dfrac{ x}{1 + x\tan x} = \dfrac{x \cos x}{\cos x + x \sin x} = \dfrac{ (\cos x + x \sin x)'}{\cos x + x \sin x} = \dfrac d{dx} \ln (\cos x + x \sin x) \\ &&\Rightarrow \int \dfrac x{1 +x \tan x} \, dx = \ln (\cos x + x \sin x) + C \end{eqnarray} \]

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How directly you had taken the differentiation on numerator ?

Can you rephrase your question? I don't understand what you're saying.

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TopNewest\[\begin{eqnarray}&& \dfrac{ x}{1 + x\tan x} = \dfrac{x \cos x}{\cos x + x \sin x} = \dfrac{ (\cos x + x \sin x)'}{\cos x + x \sin x} = \dfrac d{dx} \ln (\cos x + x \sin x) \\ &&\Rightarrow \int \dfrac x{1 +x \tan x} \, dx = \ln (\cos x + x \sin x) + C \end{eqnarray} \]

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How directly you had taken the differentiation on numerator ?

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Can you rephrase your question? I don't understand what you're saying.

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