What is the integral of 1/((4-x^2)^2) ?

The integral of 1/((4-x^2)^2) is 1/32 ln|x+2|-1/(16(x+2))-1/32 ln|x-2|-1/(16(x-2))+C

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integral of 1/((4-x^2)^2)

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

\frac{1}{32} ln ∣ x + 2 ∣ - \frac{1}{16 ( x + 2 )} - \frac{1}{32} ln ∣ x - 2 ∣ - \frac{1}{16 ( x - 2 )} + C

Solution steps

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

Take the partial fraction of

\frac{1}{( 4 - x ^{2} ) ^{2}} : \frac{1}{32 ( x + 2 )} + \frac{1}{16 ( x + 2 ) ^{2}} - \frac{1}{32 ( x - 2 )} + \frac{1}{16 ( x - 2 ) ^{2}}

= \int \frac{1}{32 ( x + 2 )} + \frac{1}{16 ( x + 2 ) ^{2}} - \frac{1}{32 ( x - 2 )} + \frac{1}{16 ( x - 2 ) ^{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

= \int \frac{1}{32 ( x + 2 )} d x + \int \frac{1}{16 ( x + 2 ) ^{2}} d x - \int \frac{1}{32 ( x - 2 )} d x + \int \frac{1}{16 ( x - 2 ) ^{2}} d x

\int \frac{1}{32 ( x + 2 )} d x = \frac{1}{32} ln ∣ x + 2 ∣

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

\int \frac{1}{32 ( x - 2 )} d x = \frac{1}{32} ln ∣ x - 2 ∣

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

= \frac{1}{32} ln ∣ x + 2 ∣ - \frac{1}{16 ( x + 2 )} - \frac{1}{32} ln ∣ x - 2 ∣ - \frac{1}{16 ( x - 2 )}

Add a constant to the solution

= \frac{1}{32} ln ∣ x + 2 ∣ - \frac{1}{16 ( x + 2 )} - \frac{1}{32} ln ∣ x - 2 ∣ - \frac{1}{16 ( x - 2 )} + C

\frac{\partial}{\partial x} (5 ln (1 + x^{2} - y))

x \to 0 lim (\frac{ln ( x )}{x ^{2}})

\int \frac{1}{1 + x + x ^{2}} d x

f^{^{'}} (x) = 5 (1 + e^{x})^{4} e^{x}

h (w) = \frac{5 w ^{3} - w}{w}

What is the integral of 1/((4-x^2)^2) ?
The integral of 1/((4-x^2)^2) is 1/32 ln|x+2|-1/(16(x+2))-1/32 ln|x-2|-1/(16(x-2))+C