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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

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

Simplify

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

Add a constant to the solution

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

cos^{3} (12 x)

y = 3 x^{2} + 2 x, (2, 4)

\int sin^{2} (x co) s^{2} x d x

y^{^{''}} - 6 y^{^{'}} + 9 y = 0, y (0) = 6, y^{^{'}} (0) = 0

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