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

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How directly you had taken the differentiation on numerator ?
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Ganesh Gamit
·
11 months, 4 weeks ago

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@Ganesh Gamit
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Can you rephrase your question? I don't understand what you're saying.
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Pi Han Goh
·
11 months, 4 weeks ago

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

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How directly you had taken the differentiation on numerator ? – Ganesh Gamit · 11 months, 4 weeks ago

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– Pi Han Goh · 11 months, 4 weeks ago

Can you rephrase your question? I don't understand what you're saying.Log in to reply