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

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

\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

x^{19} e^{x}

\frac{\partial}{\partial x} (x^{2} - 1)

\frac{d}{d x} ((x^{2} + 5 x) e^{x})

\int x^{2} y d y

\int 4 tan (t) d t

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