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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 1 to 3 of (2x-1)/((x+1)^2)

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

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

Decimal Notation

0.63629 \dots

Solution steps

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

Expand

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

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

\int_{1}^{3} \frac{2 x}{( x + 1 ) ^{2}} d x = 2 (ln (2) - \frac{1}{4})

\int_{1}^{3} \frac{1}{( x + 1 ) ^{2}} d x = \frac{1}{4}

= 2 (ln (2) - \frac{1}{4}) - \frac{1}{4}

Simplify

2 (ln (2) - \frac{1}{4}) - \frac{1}{4} : 2 ln (2) - \frac{3}{4}

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

\int_{2}^{4} (\frac{x}{2} - \frac{2 x ^{2}}{3}) d x

\int_{- 1}^{1} x^{- 1} d x

\int_{0}^{π} x^{4} cos (2 x^{5}) d x

\int_{1}^{2} y^{2} d y

\int_{3}^{6} 2 d x

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