What is the integral from 1 to 2 of 25(x^2+1)^7 ?

The integral from 1 to 2 of 25(x^2+1)^7 is 427048150/1287

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integral from 1 to 2 of 25(x^2+1)^7

\int_{1}^{2} 25 (x^{2} + 1)^{7} d x

\frac{427048150}{1287}

Decimal

331816.74436 \dots

Solution steps

\int_{1}^{2} 25 (x^{2} + 1)^{7} d x

Take the constant out:

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

= 25 \cdot \int_{1}^{2} (x^{2} + 1)^{7} d x

Expand

(x^{2} + 1)^{7} : x^{14} + 7 x^{12} + 21 x^{10} + 35 x^{8} + 35 x^{6} + 21 x^{4} + 7 x^{2} + 1

= 25 \cdot \int_{1}^{2} x^{14} + 7 x^{12} + 21 x^{10} + 35 x^{8} + 35 x^{6} + 21 x^{4} + 7 x^{2} + 1 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

= 25 (\int_{1}^{2} x^{14} d x + \int_{1}^{2} 7 x^{12} d x + \int_{1}^{2} 21 x^{10} d x + \int_{1}^{2} 35 x^{8} d x + \int_{1}^{2} 35 x^{6} d x + \int_{1}^{2} 21 x^{4} d x + \int_{1}^{2} 7 x^{2} d x + \int_{1}^{2} 1 d x)

\int_{1}^{2} x^{14} d x = \frac{32767}{15}

\int_{1}^{2} 7 x^{12} d x = \frac{57337}{13}

\int_{1}^{2} 21 x^{10} d x = \frac{42987}{11}

\int_{1}^{2} 35 x^{8} d x = \frac{17885}{9}

\int_{1}^{2} 35 x^{6} d x = 635

\int_{1}^{2} 21 x^{4} d x = \frac{651}{5}

\int_{1}^{2} 7 x^{2} d x = \frac{49}{3}

\int_{1}^{2} 1 d x = 1

= 25 (\frac{32767}{15} + \frac{57337}{13} + \frac{42987}{11} + \frac{17885}{9} + 635 + \frac{651}{5} + \frac{49}{3} + 1)

Simplify

25 (\frac{32767}{15} + \frac{57337}{13} + \frac{42987}{11} + \frac{17885}{9} + 635 + \frac{651}{5} + \frac{49}{3} + 1) : \frac{427048150}{1287}

= \frac{427048150}{1287}

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What is the integral from 1 to 2 of 25(x^2+1)^7 ?
The integral from 1 to 2 of 25(x^2+1)^7 is 427048150/1287