$\int\frac{\log(x+1)-\log x}{x(x+1)}dx$

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Target Exam

CUET

Subject

-- Mathematics - Section A

Chapter

Indefinite Integration

Question:

$\int\frac{\log(x+1)-\log x}{x(x+1)}dx$ is equal to:

Options:

$-\frac{1}{2}[\log(\frac{x+1}{x})]^2+C$

$C-[\{\log(x+1)\}^2-(\log x)^2]$

$-\log[\log(\frac{x+1}{x})]+C$

$-\log(\frac{x+1}{x})+C$

Correct Answer:

$-\frac{1}{2}[\log(\frac{x+1}{x})]^2+C$

Explanation:

$\int\frac{\log(x+1)-\log x}{x(x+1)}dx$

$\int(\log(x+1)-\log x).\frac{1}{x(x+1)}dx=\frac{-1}{2}[\log(\frac{x+1}{x})]^2+C$

$[\frac{d}{dx}[(x-1)-\log x]=\frac{1}{x+1}-\frac{1}{x}=-\frac{1}{(x+1)x}]$