\(\int \frac{\log x}{(1+\log x)^{2}}dx=\

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Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Indefinite Integration

Question:

$\int \frac{\log x}{(1+\log x)^{2}}dx=$

Options:

$\frac{x}{1-\log x}+k$

$\frac{x}{1+\log x}+k$

$\frac{1+\log x}{2}+k$

$\frac{x}{1+\log x}-k$

Correct Answer:

$\frac{x}{1+\log x}+k$

Explanation:

$x=e^y⇒dx=e^ydy$

$I=\int\frac{\log xdx}{(1+\log x)^2}=\int\frac{ye^ydy}{(1+y)^2}=\int e^y\frac{y+1-1}{(y+1)^2}dy$

$=\int e^y\left(\frac{1}{y+1}-\frac{1}{(y+1)^2}\right)dy=\frac{e^y}{y+1}+C$

$=\frac{x}{\log x+1}+C$