What is the integral of 4/(y^2+5y-14) ?

The integral of 4/(y^2+5y-14) is 4(1/9 ln|y-2|-1/9 ln|y+7|)+C

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integral of 4/(y^2+5y-14)

\int \frac{4}{y ^{2} + 5 y - 14} d y

4 (\frac{1}{9} ln ∣ y - 2 ∣ - \frac{1}{9} ln ∣ y + 7 ∣) + C

Solution steps

\int \frac{4}{y ^{2} + 5 y - 14} d y

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 4 \cdot \int \frac{1}{y ^{2} + 5 y - 14} d y

Take the partial fraction of

\frac{1}{y ^{2} + 5 y - 14} : \frac{1}{9 ( y - 2 )} - \frac{1}{9 ( y + 7 )}

= 4 \cdot \int \frac{1}{9 ( y - 2 )} - \frac{1}{9 ( y + 7 )} 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

= 4 (\int \frac{1}{9 ( y - 2 )} d y - \int \frac{1}{9 ( y + 7 )} d y)

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

\int \frac{1}{9 ( y + 7 )} d y = \frac{1}{9} ln ∣ y + 7 ∣

= 4 (\frac{1}{9} ln ∣ y - 2 ∣ - \frac{1}{9} ln ∣ y + 7 ∣)

Add a constant to the solution

= 4 (\frac{1}{9} ln ∣ y - 2 ∣ - \frac{1}{9} ln ∣ y + 7 ∣) + C

\int_{0}^{21} - x (x - 21) (x + 1) d x

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

y^{^{'}} + \frac{y}{10} = 500

\frac{d N}{d t} = N (1 - 0.0002 N)

\frac{d}{d x} (c_{1} cos (2 x) + c_{2} sin (2 x))

What is the integral of 4/(y^2+5y-14) ?
The integral of 4/(y^2+5y-14) is 4(1/9 ln|y-2|-1/9 ln|y+7|)+C