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

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

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

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

\frac{x ^{5}}{5} - x + C

Solution steps

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

Multiply by the conjugate of

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

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

Simplify

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

= \int x^{4} - 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^{4} d x - \int 1 d x

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

\int 1 d x = x

= \frac{x ^{5}}{5} - x

Add a constant to the solution

= \frac{x ^{5}}{5} - x + C

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