What is the integral of ((x-3))/((x+4)(x-5)) ?

The integral of ((x-3))/((x+4)(x-5)) is 7/9 ln|x+4|+2/9 ln|x-5|+C

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integral of ((x-3))/((x+4)(x-5))

\int \frac{( x - 3 )}{( x + 4 ) ( x - 5 )} d x

\frac{7}{9} ln ∣ x + 4 ∣ + \frac{2}{9} ln ∣ x - 5 ∣ + C

Solution steps

\int \frac{x - 3}{( x + 4 ) ( x - 5 )} d x

Take the partial fraction of

\frac{x - 3}{( x + 4 ) ( x - 5 )} : \frac{7}{9 ( x + 4 )} + \frac{2}{9 ( x - 5 )}

= \int \frac{7}{9 ( x + 4 )} + \frac{2}{9 ( x - 5 )} 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

= \int \frac{7}{9 ( x + 4 )} d x + \int \frac{2}{9 ( x - 5 )} d x

\int \frac{7}{9 ( x + 4 )} d x = \frac{7}{9} ln ∣ x + 4 ∣

\int \frac{2}{9 ( x - 5 )} d x = \frac{2}{9} ln ∣ x - 5 ∣

= \frac{7}{9} ln ∣ x + 4 ∣ + \frac{2}{9} ln ∣ x - 5 ∣

Add a constant to the solution

= \frac{7}{9} ln ∣ x + 4 ∣ + \frac{2}{9} ln ∣ x - 5 ∣ + C

\frac{d}{d x} (x^{3} - 5 x^{2} + 7 x - 3)

\frac{\partial}{\partial x} (x^{2} y - \frac{1}{x y})

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

f (x) = x^{2} 9 - x^{2}

- \frac{1}{( 1 + t ) ^{2}}

What is the integral of ((x-3))/((x+4)(x-5)) ?
The integral of ((x-3))/((x+4)(x-5)) is 7/9 ln|x+4|+2/9 ln|x-5|+C