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

The integral of 5/((x^2-1)^2) is 5(1/4 ln|x+1|-1/(4(x+1))-1/4 ln|x-1|-1/(4(x-1)))+C

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

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

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

Solution steps

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

Take the constant out:

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

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

(x^{2} - 1)^{2} = x^{4} - 2 x^{2} + 1

= 5 \cdot \int \frac{1}{x ^{4} - 2 x ^{2} + 1} d x

Take the partial fraction of

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

= 5 \cdot \int \frac{1}{4 ( x + 1 )} + \frac{1}{4 ( x + 1 ) ^{2}} - \frac{1}{4 ( x - 1 )} + \frac{1}{4 ( x - 1 ) ^{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

= 5 (\int \frac{1}{4 ( x + 1 )} d x + \int \frac{1}{4 ( x + 1 ) ^{2}} d x - \int \frac{1}{4 ( x - 1 )} d x + \int \frac{1}{4 ( x - 1 ) ^{2}} d x)

\int \frac{1}{4 ( x + 1 )} d x = \frac{1}{4} ln ∣ x + 1 ∣

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

\int \frac{1}{4 ( x - 1 )} d x = \frac{1}{4} ln ∣ x - 1 ∣

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

= 5 (\frac{1}{4} ln ∣ x + 1 ∣ - \frac{1}{4 ( x + 1 )} - \frac{1}{4} ln ∣ x - 1 ∣ - \frac{1}{4 ( x - 1 )})

Add a constant to the solution

= 5 (\frac{1}{4} ln ∣ x + 1 ∣ - \frac{1}{4 ( x + 1 )} - \frac{1}{4} ln ∣ x - 1 ∣ - \frac{1}{4 ( x - 1 )}) + C

\frac{6}{x ^{4}}

\int 2 t^{3} - 6 t + 7 d t

\frac{d}{d x} (\frac{14 x + 6 x ^{2}}{( 7 + 6 x ) ^{2}})

\frac{\partial}{\partial x} (\frac{- x - 7 y}{y ^{2} + 6 x ^{2}})

\frac{d}{d x} ((3 x + 2 y)^{5})

What is the integral of 5/((x^2-1)^2) ?
The integral of 5/((x^2-1)^2) is 5(1/4 ln|x+1|-1/(4(x+1))-1/4 ln|x-1|-1/(4(x-1)))+C