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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

= 2 \cdot \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

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

Apply the Power Rule

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

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

Substitute back

u = x^{2} - 1

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

Simplify

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

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

Add a constant to the solution

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

t \to - 2 lim (8 + t^{3})

cos (arcsin (x))

y^{^{'}} = (t + y) 2 - 1, y (1) = 6

\frac{2}{x ^{3}}

f (x) = 2 sin^{2} (π x), at x = \frac{3}{4}

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