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

The integral of y/((y+2)(4y-1)) is 2/9 ln|y+2|+1/36 ln|4y-1|+C

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

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

\frac{2}{9} ln ∣ y + 2 ∣ + \frac{1}{36} ln ∣ 4 y - 1 ∣ + C

Solution steps

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

Take the partial fraction of

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

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

\int \frac{2}{9 ( y + 2 )} d y = \frac{2}{9} ln ∣ y + 2 ∣

\int \frac{1}{9 ( 4 y - 1 )} d y = \frac{1}{36} ln ∣ 4 y - 1 ∣

= \frac{2}{9} ln ∣ y + 2 ∣ + \frac{1}{36} ln ∣ 4 y - 1 ∣

Add a constant to the solution

= \frac{2}{9} ln ∣ y + 2 ∣ + \frac{1}{36} ln ∣ 4 y - 1 ∣ + C

y = 4.5 x^{2} - 3 x + 2

\frac{\partial}{\partial x} (e^{\frac{x}{2}})

y^{^{'}} = sec^{2} (x) (1 - 3 y), y (\frac{π}{4}) = 0

f (x) = x^{2} + 16, at x = 3

\int sin^{3} (5 x) cot (5 x) d x

What is the integral of y/((y+2)(4y-1)) ?
The integral of y/((y+2)(4y-1)) is 2/9 ln|y+2|+1/36 ln|4y-1|+C