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

The integral of 1/((x+2)^2(x-5)) is -1/49 ln|x+2|+1/(7(x+2))+1/49 ln|x-5|+C

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

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

- \frac{1}{49} ln ∣ x + 2 ∣ + \frac{1}{7 ( x + 2 )} + \frac{1}{49} ln ∣ x - 5 ∣ + C

Solution steps

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

Take the partial fraction of

\frac{1}{( x + 2 ) ^{2} ( x - 5 )} : - \frac{1}{49 ( x + 2 )} - \frac{1}{7 ( x + 2 ) ^{2}} + \frac{1}{49 ( x - 5 )}

= \int - \frac{1}{49 ( x + 2 )} - \frac{1}{7 ( x + 2 ) ^{2}} + \frac{1}{49 ( 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}{49 ( x + 2 )} d x - \int \frac{1}{7 ( x + 2 ) ^{2}} d x + \int \frac{1}{49 ( x - 5 )} d x

\int \frac{1}{49 ( x + 2 )} d x = \frac{1}{49} ln ∣ x + 2 ∣

\int \frac{1}{7 ( x + 2 ) ^{2}} d x = - \frac{1}{7 ( x + 2 )}

\int \frac{1}{49 ( x - 5 )} d x = \frac{1}{49} ln ∣ x - 5 ∣

= - \frac{1}{49} ln ∣ x + 2 ∣ - (- \frac{1}{7 ( x + 2 )}) + \frac{1}{49} ln ∣ x - 5 ∣

Simplify

= - \frac{1}{49} ln ∣ x + 2 ∣ + \frac{1}{7 ( x + 2 )} + \frac{1}{49} ln ∣ x - 5 ∣

Add a constant to the solution

= - \frac{1}{49} ln ∣ x + 2 ∣ + \frac{1}{7 ( x + 2 )} + \frac{1}{49} ln ∣ x - 5 ∣ + C

\int sin (3 θ) d θ

f (x) = e^{2 x}, at x = \frac{1}{2} ln (3)

(\frac{t ^{3}}{t ^{6} + 7})^{2}

\int \frac{x}{16 + x ^{2}} d x

x \to 3 lim (2 x + ∣ x - 3 ∣)

What is the integral of 1/((x+2)^2(x-5)) ?
The integral of 1/((x+2)^2(x-5)) is -1/49 ln|x+2|+1/(7(x+2))+1/49 ln|x-5|+C