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

The integral of 2/((x+2)^2(2-x)) is 2(1/16 ln|x+2|-1/(4(x+2))-1/16 ln|x-2|)+C

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

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

2 (\frac{1}{16} ln ∣ x + 2 ∣ - \frac{1}{4 ( x + 2 )} - \frac{1}{16} ln ∣ x - 2 ∣) + C

Solution steps

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

Take the constant out:

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

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

Take the partial fraction of

\frac{1}{( x + 2 ) ^{2} ( 2 - x )} : \frac{1}{16 ( x + 2 )} + \frac{1}{4 ( x + 2 ) ^{2}} - \frac{1}{16 ( x - 2 )}

= 2 \cdot \int \frac{1}{16 ( x + 2 )} + \frac{1}{4 ( x + 2 ) ^{2}} - \frac{1}{16 ( 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

= 2 (\int \frac{1}{16 ( x + 2 )} d x + \int \frac{1}{4 ( x + 2 ) ^{2}} d x - \int \frac{1}{16 ( x - 2 )} d x)

\int \frac{1}{16 ( x + 2 )} d x = \frac{1}{16} ln ∣ x + 2 ∣

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

\int \frac{1}{16 ( x - 2 )} d x = \frac{1}{16} ln ∣ x - 2 ∣

= 2 (\frac{1}{16} ln ∣ x + 2 ∣ - \frac{1}{4 ( x + 2 )} - \frac{1}{16} ln ∣ x - 2 ∣)

Add a constant to the solution

= 2 (\frac{1}{16} ln ∣ x + 2 ∣ - \frac{1}{4 ( x + 2 )} - \frac{1}{16} ln ∣ x - 2 ∣) + C

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What is the integral of 2/((x+2)^2(2-x)) ?
The integral of 2/((x+2)^2(2-x)) is 2(1/16 ln|x+2|-1/(4(x+2))-1/16 ln|x-2|)+C