What is the integral of ln(1+1/x)x^{-2} ?

The integral of ln(1+1/x)x^{-2} is -(ln(1+1/x))/x+ln|x|+1/x-ln|x+1|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of ln(1+1/x)x^{-2}

\int ln (1 + \frac{1}{x}) x^{- 2} d x

- \frac{ln ( 1 + \frac{1}{x} )}{x} + ln ∣ x ∣ + \frac{1}{x} - ln ∣ x + 1 ∣ + C

Solution steps

\int ln (1 + \frac{1}{x}) x^{- 2} d x

Apply Integration By Parts

: - \frac{ln ( 1 + \frac{1}{x} )}{x} - \int \frac{1}{x ^{2} ( x + 1 )} d x

= - \frac{ln ( 1 + \frac{1}{x} )}{x} - \int \frac{1}{x ^{2} ( x + 1 )} d x

\int \frac{1}{x ^{2} ( x + 1 )} d x = - ln ∣ x ∣ - \frac{1}{x} + ln ∣ x + 1 ∣

= - \frac{ln ( 1 + \frac{1}{x} )}{x} - (- ln ∣ x ∣ - \frac{1}{x} + ln ∣ x + 1 ∣)

Simplify

= - \frac{ln ( 1 + \frac{1}{x} )}{x} + ln ∣ x ∣ + \frac{1}{x} - ln ∣ x + 1 ∣

Add a constant to the solution

= - \frac{ln ( 1 + \frac{1}{x} )}{x} + ln ∣ x ∣ + \frac{1}{x} - ln ∣ x + 1 ∣ + C

\int \frac{1}{sin ( θ ) + 1} d θ

\int \frac{1}{x ( x + 3 ) ^{2}} d x

\int (9 x + 8)^{2} d x

\int (x + 7) (x - 2) d x

\int \frac{sin ( 2 x ) - cos ( 2 x )}{sin ( 2 x )} d x

What is the integral of ln(1+1/x)x^{-2} ?
The integral of ln(1+1/x)x^{-2} is -(ln(1+1/x))/x+ln|x|+1/x-ln|x+1|+C