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

The integral from 1 to 2 of (x^2+5)/(3x-x^2) is -1+19/3 ln(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+5)/(3x-x^2)

\int_{1}^{2} \frac{x ^{2} + 5}{3 x - x ^{2}} d x

- 1 + \frac{19}{3} ln (2)

Decimal Notation

3.38993 \dots

Solution steps

\int_{1}^{2} \frac{x ^{2} + 5}{3 x - x ^{2}} d x

Factor

\frac{x ^{2} + 5}{3 x - x ^{2}}

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Expand

\frac{x ^{2} + 5}{x ( x - 3 )} : \frac{x ^{2}}{x ( x - 3 )} + \frac{5}{x ( x - 3 )}

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

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

\int_{1}^{2} \frac{5}{x ( x - 3 )} d x = - \frac{10}{3} ln (2)

= - (1 - 3 ln (2) - \frac{10}{3} ln (2))

Simplify

- (1 - 3 ln (2) - \frac{10}{3} ln (2)) : - 1 + \frac{19}{3} ln (2)

= - 1 + \frac{19}{3} ln (2)

\frac{d}{d x} (\frac{x ^{2}}{4 + x})

\int_{0}^{\frac{3}{4}} 60 x d x

y^{^{'}} - 3 y + 2 x = 0

x \to \infty lim (x^{2} - 4)

\int_{0}^{1} \frac{y}{e ^{7 y}} d y

What is the integral from 1 to 2 of (x^2+5)/(3x-x^2) ?
The integral from 1 to 2 of (x^2+5)/(3x-x^2) is -1+19/3 ln(2)