What is the integral of x(x^2-1)^5 ?

The integral of x(x^2-1)^5 is 1/12 (x^2-1)^6+C

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integral of x(x^2-1)^5

\int x (x^{2} - 1)^{5} d x

\frac{1}{12} (x^{2} - 1)^{6} + C

Solution steps

\int x (x^{2} - 1)^{5} d x

Apply u-substitution

: \int \frac{u ^{5}}{2} d u

= \int \frac{u ^{5}}{2} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int u^{5} d u

Apply the Power Rule

: \frac{u ^{6}}{6}

= \frac{1}{2} \cdot \frac{u ^{6}}{6}

Substitute back

u = x^{2} - 1

= \frac{1}{2} \cdot \frac{( x ^{2} - 1 ) ^{6}}{6}

Simplify

\frac{1}{2} \cdot \frac{( x ^{2} - 1 ) ^{6}}{6} : \frac{1}{12} (x^{2} - 1)^{6}

= \frac{1}{12} (x^{2} - 1)^{6}

Add a constant to the solution

= \frac{1}{12} (x^{2} - 1)^{6} + C

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

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

y^{^{''}} - 4 y^{^{'}} = e^{2 x}

(\frac{h}{2})^{2}

\int_{0}^{4} (64 x - 8 x^{2}) d x

What is the integral of x(x^2-1)^5 ?
The integral of x(x^2-1)^5 is 1/12 (x^2-1)^6+C