What is the integral of 1/(x^2(1+x)) ?

The integral of 1/(x^2(1+x)) is -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 1/(x^2(1+x))

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

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

Solution steps

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

Take the partial fraction of

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

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int \frac{1}{x} d x = ln ∣ x ∣

\int \frac{1}{x ^{2}} d x = - \frac{1}{x}

\int \frac{1}{x + 1} d x = ln ∣ x + 1 ∣

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

Add a constant to the solution

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

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

\int (3 - 2 x) cos (5 - 4 x) d x

\int 2^{5 t^{2}} t d t

\int x (ln (x) + 1) d x

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

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