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

The integral of x(x^4-1)^2 is (x^{10}}{10}-\frac{x^6)/3+(x^2)/2+C

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

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

\frac{x ^{10}}{10} - \frac{x ^{6}}{3} + \frac{x ^{2}}{2} + C

Solution steps

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

Expand

x (x^{4} - 1)^{2} : x^{9} - 2 x^{5} + x

= \int x^{9} - 2 x^{5} + x 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^{9} d x - \int 2 x^{5} d x + \int x d x

\int x^{9} d x = \frac{x ^{10}}{10}

\int 2 x^{5} d x = \frac{x ^{6}}{3}

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

= \frac{x ^{10}}{10} - \frac{x ^{6}}{3} + \frac{x ^{2}}{2}

Add a constant to the solution

= \frac{x ^{10}}{10} - \frac{x ^{6}}{3} + \frac{x ^{2}}{2} + C

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

x \to - 2 π lim (sin (x))

h \to 1 lim (\frac{( h - 1 ) ^{3} + 1}{h})

x \to 0 lim (\frac{arcsin ( x )}{x})

\frac{\partial}{\partial y} (sin (y) - tan (x) - y)

What is the integral of x(x^4-1)^2 ?
The integral of x(x^4-1)^2 is (x^{10}}{10}-\frac{x^6)/3+(x^2)/2+C