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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

Take the partial fraction of

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

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

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 ( x - 2 )} d x - \int \frac{1}{9 ( x + 7 )} d x)

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

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

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

Add a constant to the solution

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

\int ln (14 x) d x

x \to 1 lim (cos (π x))

\int \frac{cos ( 2 x )}{sin ( 2 x )} d x

x \to 8 lim (\frac{x ^{2}}{4} - \frac{2}{x})

\frac{d}{d x} (\frac{x ^{2} + 4 x + 3}{x})

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