What is the integral of (28)/(x^2-4x-45) ?

The integral of (28)/(x^2-4x-45) is 28(-1/14 ln|x+5|+1/14 ln|x-9|)+C

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integral of (28)/(x^2-4x-45)

\int \frac{28}{x ^{2} - 4 x - 45} d x

28 (- \frac{1}{14} ln ∣ x + 5 ∣ + \frac{1}{14} ln ∣ x - 9 ∣) + C

Solution steps

\int \frac{28}{x ^{2} - 4 x - 45} d x

Take the constant out:

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

= 28 \cdot \int \frac{1}{x ^{2} - 4 x - 45} d x

Take the partial fraction of

\frac{1}{x ^{2} - 4 x - 45} : - \frac{1}{14 ( x + 5 )} + \frac{1}{14 ( x - 9 )}

= 28 \cdot \int - \frac{1}{14 ( x + 5 )} + \frac{1}{14 ( x - 9 )} 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

= 28 (- \int \frac{1}{14 ( x + 5 )} d x + \int \frac{1}{14 ( x - 9 )} d x)

\int \frac{1}{14 ( x + 5 )} d x = \frac{1}{14} ln ∣ x + 5 ∣

\int \frac{1}{14 ( x - 9 )} d x = \frac{1}{14} ln ∣ x - 9 ∣

= 28 (- \frac{1}{14} ln ∣ x + 5 ∣ + \frac{1}{14} ln ∣ x - 9 ∣)

Add a constant to the solution

= 28 (- \frac{1}{14} ln ∣ x + 5 ∣ + \frac{1}{14} ln ∣ x - 9 ∣) + C

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What is the integral of (28)/(x^2-4x-45) ?
The integral of (28)/(x^2-4x-45) is 28(-1/14 ln|x+5|+1/14 ln|x-9|)+C