What is the integral of y/((y+4)(2y-1)) ?

The integral of y/((y+4)(2y-1)) is 4/9 ln|y+4|+1/18 ln|2y-1|+C

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integral of y/((y+4)(2y-1))

\int \frac{y}{( y + 4 ) ( 2 y - 1 )} d y

\frac{4}{9} ln ∣ y + 4 ∣ + \frac{1}{18} ln ∣ 2 y - 1 ∣ + C

Solution steps

\int \frac{y}{( y + 4 ) ( 2 y - 1 )} d y

Take the partial fraction of

\frac{y}{( y + 4 ) ( 2 y - 1 )} : \frac{4}{9 ( y + 4 )} + \frac{1}{9 ( 2 y - 1 )}

= \int \frac{4}{9 ( y + 4 )} + \frac{1}{9 ( 2 y - 1 )} 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{4}{9 ( y + 4 )} d y + \int \frac{1}{9 ( 2 y - 1 )} d y

\int \frac{4}{9 ( y + 4 )} d y = \frac{4}{9} ln ∣ y + 4 ∣

\int \frac{1}{9 ( 2 y - 1 )} d y = \frac{1}{18} ln ∣ 2 y - 1 ∣

= \frac{4}{9} ln ∣ y + 4 ∣ + \frac{1}{18} ln ∣ 2 y - 1 ∣

Add a constant to the solution

= \frac{4}{9} ln ∣ y + 4 ∣ + \frac{1}{18} ln ∣ 2 y - 1 ∣ + C

\frac{d}{d x} (- x^{\frac{1}{2}})

\int e^{x + 2 y} d y

\frac{d ^{3}}{d x ^{3}} (\frac{1}{x})

t (\frac{d y}{d t}) + 22 y = \frac{1}{t}

\frac{\partial}{\partial y} (y ln (y))

What is the integral of y/((y+4)(2y-1)) ?
The integral of y/((y+4)(2y-1)) is 4/9 ln|y+4|+1/18 ln|2y-1|+C