What is the integral of ((3y-8))/(y^2-4y-5) ?

The integral of ((3y-8))/(y^2-4y-5) is 11/6 ln|y+1|+7/6 ln|y-5|+C

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integral of ((3y-8))/(y^2-4y-5)

\int \frac{( 3 y - 8 )}{y ^{2} - 4 y - 5} d y

\frac{11}{6} ln ∣ y + 1 ∣ + \frac{7}{6} ln ∣ y - 5 ∣ + C

Solution steps

\int \frac{3 y - 8}{y ^{2} - 4 y - 5} d y

Take the partial fraction of

\frac{3 y - 8}{y ^{2} - 4 y - 5} : \frac{11}{6 ( y + 1 )} + \frac{7}{6 ( y - 5 )}

= \int \frac{11}{6 ( y + 1 )} + \frac{7}{6 ( y - 5 )} 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 \frac{11}{6 ( y + 1 )} d y + \int \frac{7}{6 ( y - 5 )} d y

\int \frac{11}{6 ( y + 1 )} d y = \frac{11}{6} ln ∣ y + 1 ∣

\int \frac{7}{6 ( y - 5 )} d y = \frac{7}{6} ln ∣ y - 5 ∣

= \frac{11}{6} ln ∣ y + 1 ∣ + \frac{7}{6} ln ∣ y - 5 ∣

Add a constant to the solution

= \frac{11}{6} ln ∣ y + 1 ∣ + \frac{7}{6} ln ∣ y - 5 ∣ + C

\int \frac{1}{( 3 x + 7 ) ^{2}} d x

\frac{s ^{2}}{s - 1}

e^{- 13 x}

\frac{d y}{d x} = \frac{( x ^{2} - 3 y ^{2} )}{2 x y}

\frac{\partial}{\partial x} (\frac{4 s + t}{3 s - 5 t})

What is the integral of ((3y-8))/(y^2-4y-5) ?
The integral of ((3y-8))/(y^2-4y-5) is 11/6 ln|y+1|+7/6 ln|y-5|+C