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

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

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integral of 1/((t^2-1)^2)

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

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

Solution steps

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

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

= \int \frac{1}{t ^{4} - 2 t ^{2} + 1} d t

Take the partial fraction of

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

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

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

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

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

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

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

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

Add a constant to the solution

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

n = 1 \sum \infty \frac{4}{4 n ^{2} - 1}

\frac{\partial}{\partial x} (x^{2} y z^{2} - 4 cos (3 yz))

f (x) = - 3 x^{2} + 1 x, at x = 0

\frac{d}{d x} (1 + 3 sin^{3} (x))

x^{2} + 1 + x^{2}

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