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

The integral of (x^2-1)^2 is (x^5)/5-(2x^3)/3+x+C

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integral of (x^2-1)^2

integral

(x^{2} - 1)^{2}

\frac{x ^{5}}{5} - \frac{2 x ^{3}}{3} + x + C

Solution steps

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

Expand

(x^{2} - 1)^{2} : x^{4} - 2 x^{2} + 1

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

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

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

\int 1 d x = x

= \frac{x ^{5}}{5} - \frac{2 x ^{3}}{3} + x

Add a constant to the solution

= \frac{x ^{5}}{5} - \frac{2 x ^{3}}{3} + x + C

(1 - 4 x) (1 + x)

3 - 2 (x)^{- 1}

\frac{1}{( x )}

\frac{x ^{2}}{( 4 + x ^{6} )}

\frac{1}{x} + \frac{1}{x ^{3}}

What is the integral of (x^2-1)^2 ?
The integral of (x^2-1)^2 is (x^5)/5-(2x^3)/3+x+C