What is the integral from 1 to 3 of (x^2+8)/(4x-x^2) ?

The integral from 1 to 3 of (x^2+8)/(4x-x^2) is -2+8ln(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 3 of (x^2+8)/(4x-x^2)

\int_{1}^{3} \frac{x ^{2} + 8}{4 x - x ^{2}} d x

- 2 + 8 ln (3)

Decimal Notation

6.78889 \dots

Solution steps

\int_{1}^{3} \frac{x ^{2} + 8}{4 x - x ^{2}} d x

Factor

\frac{x ^{2} + 8}{4 x - x ^{2}}

= \int_{1}^{3} \frac{x ^{2} + 8}{- x ( x - 4 )} d x

Take the constant out:

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

= - \int_{1}^{3} \frac{x ^{2} + 8}{x ( x - 4 )} d x

Expand

\frac{x ^{2} + 8}{x ( x - 4 )} : \frac{x ^{2}}{x ( x - 4 )} + \frac{8}{x ( x - 4 )}

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{x ^{2}}{x ( x - 4 )} d x + \int_{1}^{3} \frac{8}{x ( x - 4 )} d x)

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

\int_{1}^{3} \frac{8}{x ( x - 4 )} d x = - 4 ln (3)

= - (2 - 4 ln (3) - 4 ln (3))

Simplify

= - 2 + 8 ln (3)

y^{^{'}} = x y

y = ln (x + x^{2} - 1)

x \to 0 + lim (ln (\frac{1}{x}))

x \to 1 lim (7 ∣ x + 1 ∣ - 4)

y = (x - 1) (x^{2} + x + 1)

What is the integral from 1 to 3 of (x^2+8)/(4x-x^2) ?
The integral from 1 to 3 of (x^2+8)/(4x-x^2) is -2+8ln(3)