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

The integral from-1 to 2 of (x^2)/(x+2) is 8ln(2)-9/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

8 ln (2) - \frac{9}{2}

Solution steps

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

Apply u-substitution

: \int_{1}^{4} \frac{( u - 2 ) ^{2}}{u} d u

= \int_{1}^{4} \frac{( u - 2 ) ^{2}}{u} d u

Expand

\frac{( u - 2 ) ^{2}}{u} : u - 4 + \frac{4}{u}

= \int_{1}^{4} u - 4 + \frac{4}{u} d u

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}^{4} u d u - \int_{1}^{4} 4 d u + \int_{1}^{4} \frac{4}{u} d u

\int_{1}^{4} u d u = \frac{15}{2}

\int_{1}^{4} 4 d u = 12

\int_{1}^{4} \frac{4}{u} d u = 8 ln (2)

= \frac{15}{2} - 12 + 8 ln (2)

Combine the fractions

- 12 + \frac{15}{2} : - \frac{9}{2}

= 8 ln (2) - \frac{9}{2}

\int_{0}^{3} 6 x - x^{2} d x

\int_{1}^{2} x^{2} + \frac{1}{x ^{4}} d x

\int_{0}^{π} x \cdot sin (x^{2}) d x

\int_{3}^{4} \frac{2}{25 - x ^{2}} d x

\int_{0}^{x} (e^{t^{2}} - 1) d t

What is the integral from-1 to 2 of (x^2)/(x+2) ?
The integral from-1 to 2 of (x^2)/(x+2) is 8ln(2)-9/2