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

The integral of x(x^2-1)^2 is (x^6)/6-(x^4)/2+(x^2)/2+C

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

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

\frac{x ^{6}}{6} - \frac{x ^{4}}{2} + \frac{x ^{2}}{2} + C

Solution steps

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

Expand

x (x^{2} - 1)^{2} : x^{5} - 2 x^{3} + x

= \int x^{5} - 2 x^{3} + 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^{5} d x - \int 2 x^{3} d x + \int x d x

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

\int 2 x^{3} d x = \frac{x ^{4}}{2}

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

= \frac{x ^{6}}{6} - \frac{x ^{4}}{2} + \frac{x ^{2}}{2}

Add a constant to the solution

= \frac{x ^{6}}{6} - \frac{x ^{4}}{2} + \frac{x ^{2}}{2} + C

(1 + x)^{n}

f (x) = - \frac{12}{( x - 6 ) ^{2}}

\int \frac{5 x ^{3} + 3 x}{x - 1} d x

\int (x^{5} + 1) d x

y^{^{''}} + 8 y^{^{'}} - 9 y = e^{3 x}

What is the integral of x(x^2-1)^2 ?
The integral of x(x^2-1)^2 is (x^6)/6-(x^4)/2+(x^2)/2+C