What is the integral from-1 to 1 of (x^2+4)(x-1)^6 ?

The integral from-1 to 1 of (x^2+4)(x-1)^6 is 5312/63

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integral from-1 to 1 of (x^2+4)(x-1)^6

\int_{- 1}^{1} (x^{2} + 4) (x - 1)^{6} d x

\frac{5312}{63}

Decimal

84.31746 \dots

Solution steps

\int_{- 1}^{1} (x^{2} + 4) (x - 1)^{6} d x

Expand

(x^{2} + 4) (x - 1)^{6} : x^{8} - 6 x^{7} + 19 x^{6} - 44 x^{5} + 75 x^{4} - 86 x^{3} + 61 x^{2} - 24 x + 4

= \int_{- 1}^{1} x^{8} - 6 x^{7} + 19 x^{6} - 44 x^{5} + 75 x^{4} - 86 x^{3} + 61 x^{2} - 24 x + 4 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_{- 1}^{1} x^{8} d x - \int_{- 1}^{1} 6 x^{7} d x + \int_{- 1}^{1} 19 x^{6} d x - \int_{- 1}^{1} 44 x^{5} d x + \int_{- 1}^{1} 75 x^{4} d x - \int_{- 1}^{1} 86 x^{3} d x + \int_{- 1}^{1} 61 x^{2} d x - \int_{- 1}^{1} 24 x d x + \int_{- 1}^{1} 4 d x

\int_{- 1}^{1} x^{8} d x = \frac{2}{9}

\int_{- 1}^{1} 6 x^{7} d x = 0

\int_{- 1}^{1} 19 x^{6} d x = \frac{38}{7}

\int_{- 1}^{1} 44 x^{5} d x = 0

\int_{- 1}^{1} 75 x^{4} d x = 30

\int_{- 1}^{1} 86 x^{3} d x = 0

\int_{- 1}^{1} 61 x^{2} d x = \frac{122}{3}

\int_{- 1}^{1} 24 x d x = 0

\int_{- 1}^{1} 4 d x = 8

= \frac{2}{9} - 0 + \frac{38}{7} - 0 + 30 - 0 + \frac{122}{3} - 0 + 8

Simplify

\frac{2}{9} - 0 + \frac{38}{7} - 0 + 30 - 0 + \frac{122}{3} - 0 + 8 : \frac{5312}{63}

= \frac{5312}{63}

What is the integral from-1 to 1 of (x^2+4)(x-1)^6 ?
The integral from-1 to 1 of (x^2+4)(x-1)^6 is 5312/63