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

The integral from 1 to 4 of (x^2+4)/(5x-x^2) is -3+66/5 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 4 of (x^2+4)/(5x-x^2)

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

- 3 + \frac{66}{5} ln (2)

Decimal Notation

6.14954 \dots

Solution steps

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

Factor

\frac{x ^{2} + 4}{5 x - x ^{2}}

= \int_{1}^{4} \frac{x ^{2} + 4}{- x ( x - 5 )} d x

Take the constant out:

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

= - \int_{1}^{4} \frac{x ^{2} + 4}{x ( x - 5 )} d x

Expand

\frac{x ^{2} + 4}{x ( x - 5 )} : \frac{x ^{2}}{x ( x - 5 )} + \frac{4}{x ( x - 5 )}

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} \frac{x ^{2}}{x ( x - 5 )} d x + \int_{1}^{4} \frac{4}{x ( x - 5 )} d x)

\int_{1}^{4} \frac{x ^{2}}{x ( x - 5 )} d x = 3 - 10 ln (2)

\int_{1}^{4} \frac{4}{x ( x - 5 )} d x = - \frac{16}{5} ln (2)

= - (3 - 10 ln (2) - \frac{16}{5} ln (2))

Simplify

- (3 - 10 ln (2) - \frac{16}{5} ln (2)) : - 3 + \frac{66}{5} ln (2)

= - 3 + \frac{66}{5} ln (2)

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

\frac{d}{d x} (\frac{d y}{d x}) + 81 y = 0

\int cos (\frac{u}{v}) d u

x \to + 0 lim (\frac{sin ( 0 )}{0})

\int 6 t^{4} e^{- t^{5}} d t

What is the integral from 1 to 4 of (x^2+4)/(5x-x^2) ?
The integral from 1 to 4 of (x^2+4)/(5x-x^2) is -3+66/5 ln(2)