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

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

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

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

arctan (t) + C

Solution steps

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

Simplify

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

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

Use the common integral:

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

= arctan (t)

Add a constant to the solution

= arctan (t) + C

y^{^{''}} + 4 y^{^{'}} + 4 y = e^{- 2 x} \cdot ln (x)

\frac{d}{d x} (\frac{sin ( x )}{b sin ( b x )})

y^{^{'}} - 2 x y = 2 x e^{x^{2}}

((4 x^{2} + 5)^{2} \cdot (7 x)^{2 x})^{^{'}}

- 4 x^{2} y + y^{^{'}} = 4 x^{2}

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