What is the integral of ((9y+14))/((y-2)(y^2+4)) ?

The integral of ((9y+14))/((y-2)(y^2+4)) is 4ln|y-2|-2ln|y^2+4|+1/2 arctan(y/2)+C

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integral of ((9y+14))/((y-2)(y^2+4))

\int \frac{( 9 y + 14 )}{( y - 2 ) ( y ^{2} + 4 )} d y

4 ln ∣ y - 2 ∣ - 2 ln y^{2} + 4 + \frac{1}{2} arctan (\frac{y}{2}) + C

Solution steps

\int \frac{9 y + 14}{( y - 2 ) ( y ^{2} + 4 )} d y

Take the partial fraction of

\frac{9 y + 14}{( y - 2 ) ( y ^{2} + 4 )} : \frac{4}{y - 2} + \frac{- 4 y + 1}{y ^{2} + 4}

= \int \frac{4}{y - 2} + \frac{- 4 y + 1}{y ^{2} + 4} 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}{y - 2} d y + \int \frac{- 4 y + 1}{y ^{2} + 4} d y

\int \frac{4}{y - 2} d y = 4 ln ∣ y - 2 ∣

\int \frac{- 4 y + 1}{y ^{2} + 4} d y = - 2 ln y^{2} + 4 + \frac{1}{2} arctan (\frac{y}{2})

= 4 ln ∣ y - 2 ∣ - 2 ln y^{2} + 4 + \frac{1}{2} arctan (\frac{y}{2})

Add a constant to the solution

= 4 ln ∣ y - 2 ∣ - 2 ln y^{2} + 4 + \frac{1}{2} arctan (\frac{y}{2}) + C

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What is the integral of ((9y+14))/((y-2)(y^2+4)) ?
The integral of ((9y+14))/((y-2)(y^2+4)) is 4ln|y-2|-2ln|y^2+4|+1/2 arctan(y/2)+C