What is the integral from 1 to 2 of (4x^2-7x-12)/(x(x+2)(x-3)) ?

The integral from 1 to 2 of (4x^2-7x-12)/(x(x+2)(x-3)) is 2ln(2)+(17ln(2)-9ln(3))/5

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 (4x^2-7x-12)/(x(x+2)(x-3))

\int_{1}^{2} \frac{4 x ^{2} - 7 x - 12}{x ( x + 2 ) ( x - 3 )} d x

2 ln (2) + \frac{17 ln ( 2 ) - 9 ln ( 3 )}{5}

Decimal Notation

1.76549 \dots

Solution steps

\int_{1}^{2} \frac{4 x ^{2} - 7 x - 12}{x ( x + 2 ) ( x - 3 )} d x

Take the partial fraction of

\frac{4 x ^{2} - 7 x - 12}{x ( x + 2 ) ( x - 3 )} : \frac{2}{x} + \frac{9}{5 ( x + 2 )} + \frac{1}{5 ( x - 3 )}

= \int_{1}^{2} \frac{2}{x} + \frac{9}{5 ( x + 2 )} + \frac{1}{5 ( x - 3 )} d x

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{2}{x} d x + \int_{1}^{2} \frac{9}{5 ( x + 2 )} d x + \int_{1}^{2} \frac{1}{5 ( x - 3 )} d x

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

\int_{1}^{2} \frac{9}{5 ( x + 2 )} d x = \frac{9}{5} (2 ln (2) - ln (3))

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

= 2 ln (2) + \frac{9}{5} (2 ln (2) - ln (3)) - \frac{1}{5} ln (2)

Simplify

2 ln (2) + \frac{9}{5} (2 ln (2) - ln (3)) - \frac{1}{5} ln (2) : 2 ln (2) + \frac{17 ln ( 2 ) - 9 ln ( 3 )}{5}

= 2 ln (2) + \frac{17 ln ( 2 ) - 9 ln ( 3 )}{5}

\frac{d}{d x} (x^{- \frac{3}{5}})

y = ln (x^{2} + 5)

\int \frac{1}{x x ^{2} - 16} d x

\frac{d y}{d x} = \frac{x - y ^{2}}{2 x y + y}

\int (sin (2 x) + cos (2 x)) d x

What is the integral from 1 to 2 of (4x^2-7x-12)/(x(x+2)(x-3)) ?
The integral from 1 to 2 of (4x^2-7x-12)/(x(x+2)(x-3)) is 2ln(2)+(17ln(2)-9ln(3))/5