What is the integral of (x^2-1)^32x ?

The integral of (x^2-1)^32x is (x^8)/4-x^6+(3x^4)/2-x^2+C

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integral of (x^2-1)^32x

\int (x^{2} - 1)^{3} 2 x d x

\frac{x ^{8}}{4} - x^{6} + \frac{3 x ^{4}}{2} - x^{2} + C

Solution steps

\int (x^{2} - 1)^{3} \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)^{3} x d x

Expand

(x^{2} - 1)^{3} x : x^{7} - 3 x^{5} + 3 x^{3} - x

= 2 \cdot \int x^{7} - 3 x^{5} + 3 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

= 2 (\int x^{7} d x - \int 3 x^{5} d x + \int 3 x^{3} d x - \int x d x)

\int x^{7} d x = \frac{x ^{8}}{8}

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

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

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

= 2 (\frac{x ^{8}}{8} - \frac{x ^{6}}{2} + \frac{3 x ^{4}}{4} - \frac{x ^{2}}{2})

Simplify

2 (\frac{x ^{8}}{8} - \frac{x ^{6}}{2} + \frac{3 x ^{4}}{4} - \frac{x ^{2}}{2}) : \frac{x ^{8}}{4} - x^{6} + \frac{3 x ^{4}}{2} - x^{2}

= \frac{x ^{8}}{4} - x^{6} + \frac{3 x ^{4}}{2} - x^{2}

Add a constant to the solution

= \frac{x ^{8}}{4} - x^{6} + \frac{3 x ^{4}}{2} - x^{2} + C

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What is the integral of (x^2-1)^32x ?
The integral of (x^2-1)^32x is (x^8)/4-x^6+(3x^4)/2-x^2+C