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

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

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

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

arctan (x) + C

Solution steps

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

Simplify

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

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

Use the common integral:

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

= arctan (x)

Add a constant to the solution

= arctan (x) + C

\int sec^{2} (x) + 4 d x

\int e^{x} e^{x} d x

\int \frac{3 x}{x ^{2} - x - 6} d x

\int_{0}^{1} 3 x (x^{2} + 1)^{4} d x

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

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