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

The integral from 4 to 5 of (4x^2+x+2)/(x-3) is 41ln(2)+31

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 4 to 5 of (4x^2+x+2)/(x-3)

\int_{4}^{5} \frac{4 x ^{2} + x + 2}{x - 3} d x

41 ln (2) + 31

Decimal

59.41903 \dots

Solution steps

\int_{4}^{5} \frac{4 x ^{2} + x + 2}{x - 3} d x

Expand

\frac{4 x ^{2} + x + 2}{x - 3} : \frac{4 x ^{2}}{x - 3} + \frac{x}{x - 3} + \frac{2}{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_{4}^{5} \frac{4 x ^{2}}{x - 3} d x + \int_{4}^{5} \frac{x}{x - 3} d x + \int_{4}^{5} \frac{2}{x - 3} d x

\int_{4}^{5} \frac{4 x ^{2}}{x - 3} d x = 4 (9 ln (2) + \frac{15}{2})

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

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

= 4 (9 ln (2) + \frac{15}{2}) + 1 + 3 ln (2) + 2 ln (2)

Simplify

4 (9 ln (2) + \frac{15}{2}) + 1 + 3 ln (2) + 2 ln (2) : 41 ln (2) + 31

= 41 ln (2) + 31

\int_{0}^{L} x^{3} d x

\int_{1}^{6} (\frac{x}{6} - \frac{5}{x}) d x

\int_{0}^{l n (3)} (2 e^{3 x}) d x

\int_{0}^{5} sin (- 1 + x) d x

\int_{o}^{π} sin (x) d x

What is the integral from 4 to 5 of (4x^2+x+2)/(x-3) ?
The integral from 4 to 5 of (4x^2+x+2)/(x-3) is 41ln(2)+31