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

The integral of (x^4-1)/(x^2+1) is (x^3)/3-x+C

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

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

\frac{x ^{3}}{3} - x + C

Solution steps

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

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

= \int x^{2} - 1 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 x^{2} d x - \int 1 d x

\int x^{2} d x = \frac{x ^{3}}{3}

\int 1 d x = x

= \frac{x ^{3}}{3} - x

Add a constant to the solution

= \frac{x ^{3}}{3} - x + C

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

x^{- 2}, 4

\int \frac{5}{4 x} d x

\int e^{- x - 1} d x

\frac{\partial}{\partial x} (sin (2 x + 5 y))

What is the integral of (x^4-1)/(x^2+1) ?
The integral of (x^4-1)/(x^2+1) is (x^3)/3-x+C