What is the integral from 1 to 5 of (x-1)/(x^2(x+1)) ?

The integral from 1 to 5 of (x-1)/(x^2(x+1)) is 2ln(5)-4/5-2ln(3)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 1 to 5 of (x-1)/(x^2(x+1))

\int_{1}^{5} \frac{x - 1}{x ^{2} ( x + 1 )} d x

2 ln (5) - \frac{4}{5} - 2 ln (3)

Decimal Notation

0.22165 \dots

Solution steps

\int_{1}^{5} \frac{x - 1}{x ^{2} ( x + 1 )} d x

Take the partial fraction of

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

= \int_{1}^{5} \frac{2}{x} - \frac{1}{x ^{2}} - \frac{2}{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_{1}^{5} \frac{2}{x} d x - \int_{1}^{5} \frac{1}{x ^{2}} d x - \int_{1}^{5} \frac{2}{x + 1} d x

\int_{1}^{5} \frac{2}{x} d x = 2 ln (5)

\int_{1}^{5} \frac{1}{x ^{2}} d x = \frac{4}{5}

\int_{1}^{5} \frac{2}{x + 1} d x = 2 ln (3)

= 2 ln (5) - \frac{4}{5} - 2 ln (3)

\int_{0}^{2} 5 x^{2} - 11 x d x

\int_{1}^{e} (22 + ln (x))^{2} d x

\int_{0}^{1000} e^{4 z} - 5 d z

\int_{0}^{4} \frac{1}{5} e^{- \frac{x}{5}} d x

\int_{- 2}^{5} (x^{2} + 6 x + 8) d x

What is the integral from 1 to 5 of (x-1)/(x^2(x+1)) ?
The integral from 1 to 5 of (x-1)/(x^2(x+1)) is 2ln(5)-4/5-2ln(3)