\[\begin{eqnarray}&& \dfrac{ x}{1 + x\tan x} = \dfrac{x \cos x}{\cos x + x \sin x} = \dfrac{ (\cos x + x \sin x)'}{\cos x + x \sin x} = \dfrac d{dx} \ln (\cos x + x \sin x) \\ &&\Rightarrow \int \dfrac x{1 +x \tan x} \, dx = \ln (\cos x + x \sin x) + C \end{eqnarray} \]
–
Pi Han Goh
·
11 months, 2 weeks ago

Log in to reply

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

Log in to reply

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

## Comments

Sort by:

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 · 11 months, 2 weeks ago

Log in to reply

How directly you had taken the differentiation on numerator ? – Ganesh Gamit · 10 months, 4 weeks ago

Log in to reply

– Pi Han Goh · 10 months, 4 weeks ago

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