What is the integral of 1/((x-5)^2(x+3)) ?

The integral of 1/((x-5)^2(x+3)) is -1/64 ln|x-5|-1/(8(x-5))+1/64 ln|x+3|+C

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integral of 1/((x-5)^2(x+3))

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

- \frac{1}{64} ln ∣ x - 5 ∣ - \frac{1}{8 ( x - 5 )} + \frac{1}{64} ln ∣ x + 3 ∣ + C

Solution steps

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

Take the partial fraction of

\frac{1}{( x - 5 ) ^{2} ( x + 3 )} : - \frac{1}{64 ( x - 5 )} + \frac{1}{8 ( x - 5 ) ^{2}} + \frac{1}{64 ( x + 3 )}

= \int - \frac{1}{64 ( x - 5 )} + \frac{1}{8 ( x - 5 ) ^{2}} + \frac{1}{64 ( x + 3 )} 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{1}{64 ( x - 5 )} d x + \int \frac{1}{8 ( x - 5 ) ^{2}} d x + \int \frac{1}{64 ( x + 3 )} d x

\int \frac{1}{64 ( x - 5 )} d x = \frac{1}{64} ln ∣ x - 5 ∣

\int \frac{1}{8 ( x - 5 ) ^{2}} d x = - \frac{1}{8 ( x - 5 )}

\int \frac{1}{64 ( x + 3 )} d x = \frac{1}{64} ln ∣ x + 3 ∣

= - \frac{1}{64} ln ∣ x - 5 ∣ - \frac{1}{8 ( x - 5 )} + \frac{1}{64} ln ∣ x + 3 ∣

Add a constant to the solution

= - \frac{1}{64} ln ∣ x - 5 ∣ - \frac{1}{8 ( x - 5 )} + \frac{1}{64} ln ∣ x + 3 ∣ + C

x \frac{d y}{d x} + y = 4 x + 1

f (x) = \frac{3}{x}, at x = 3

\frac{d}{d x} (\frac{3 x - 3}{2 x})

\frac{d y}{d x} = \frac{3}{1 + x}

f (x) = 3 cos (2 x)

What is the integral of 1/((x-5)^2(x+3)) ?
The integral of 1/((x-5)^2(x+3)) is -1/64 ln|x-5|-1/(8(x-5))+1/64 ln|x+3|+C