What is the integral of (10(x-9))/((x+5)(x-2)) ?

The integral of (10(x-9))/((x+5)(x-2)) is 10(2ln|x+5|-ln|x-2|)+C

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integral of (10(x-9))/((x+5)(x-2))

\int \frac{10 ( x - 9 )}{( x + 5 ) ( x - 2 )} d x

10 (2 ln ∣ x + 5 ∣ - ln ∣ x - 2 ∣) + C

Solution steps

\int \frac{10 ( x - 9 )}{( x + 5 ) ( x - 2 )} d x

Take the constant out:

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

= 10 \cdot \int \frac{x - 9}{( x + 5 ) ( x - 2 )} d x

Take the partial fraction of

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

= 10 \cdot \int \frac{2}{x + 5} - \frac{1}{x - 2} 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

= 10 (\int \frac{2}{x + 5} d x - \int \frac{1}{x - 2} d x)

\int \frac{2}{x + 5} d x = 2 ln ∣ x + 5 ∣

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

= 10 (2 ln ∣ x + 5 ∣ - ln ∣ x - 2 ∣)

Add a constant to the solution

= 10 (2 ln ∣ x + 5 ∣ - ln ∣ x - 2 ∣) + C

\int \frac{5 x ( 2 - x ^{2} ) ^{\frac{1}{4}}}{4} d x

\int \frac{1}{x ^{2}} + \frac{1}{x ^{3}} d x

\int \frac{x ^{2} - 5 x + 2}{x ^{3} - x ^{2} - x + 1} d x

\int - \frac{24}{x ^{25}} d x

What is the integral of (10(x-9))/((x+5)(x-2)) ?
The integral of (10(x-9))/((x+5)(x-2)) is 10(2ln|x+5|-ln|x-2|)+C