What is the integral from 1 to 2 of (4y^2-6y-12)/(y(y+2)(y-3)) ?

The integral from 1 to 2 of (4y^2-6y-12)/(y(y+2)(y-3)) is 2ln(2)+(14ln(2)-8ln(3))/5

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integral from 1 to 2 of (4y^2-6y-12)/(y(y+2)(y-3))

\int_{1}^{2} \frac{4 y ^{2} - 6 y - 12}{y ( y + 2 ) ( y - 3 )} d y

2 ln (2) + \frac{14 ln ( 2 ) - 8 ln ( 3 )}{5}

Decimal Notation

1.56932 \dots

Solution steps

\int_{1}^{2} \frac{4 y ^{2} - 6 y - 12}{y ( y + 2 ) ( y - 3 )} d y

Take the partial fraction of

\frac{4 y ^{2} - 6 y - 12}{y ( y + 2 ) ( y - 3 )} : \frac{2}{y} + \frac{8}{5 ( y + 2 )} + \frac{2}{5 ( y - 3 )}

= \int_{1}^{2} \frac{2}{y} + \frac{8}{5 ( y + 2 )} + \frac{2}{5 ( y - 3 )} d y

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}{y} d y + \int_{1}^{2} \frac{8}{5 ( y + 2 )} d y + \int_{1}^{2} \frac{2}{5 ( y - 3 )} d y

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

\int_{1}^{2} \frac{8}{5 ( y + 2 )} d y = \frac{8}{5} (2 ln (2) - ln (3))

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

= 2 ln (2) + \frac{8}{5} (2 ln (2) - ln (3)) - \frac{2}{5} ln (2)

Simplify

2 ln (2) + \frac{8}{5} (2 ln (2) - ln (3)) - \frac{2}{5} ln (2) : 2 ln (2) + \frac{14 ln ( 2 ) - 8 ln ( 3 )}{5}

= 2 ln (2) + \frac{14 ln ( 2 ) - 8 ln ( 3 )}{5}

\int x^{2} sinh (x) d x

f (x) = 2 x^{- 2} - 3 x^{- \frac{7}{3}}

\int 3 x x - 5 d x

\frac{\partial}{\partial x} (arcsin (x))

\frac{d}{d x} (4 x - 5)

What is the integral from 1 to 2 of (4y^2-6y-12)/(y(y+2)(y-3)) ?
The integral from 1 to 2 of (4y^2-6y-12)/(y(y+2)(y-3)) is 2ln(2)+(14ln(2)-8ln(3))/5