What is the integral from x to 4 of (2+sqrt(u))^8 ?

The integral from x to 4 of (2+sqrt(u))^8 is

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integral from x to 4 of (2+sqrt(u))^8

\int_{x}^{4} (2 + u)^{8} d u

- 28 x^{4} - 896 x^{2} - 256 x - 128 x^{\frac{7}{2}} + \frac{- x ^{5} + 1024}{5} + \frac{1120 ( - x ^{3} + 64 )}{3} + \frac{1024 ( - 2 x ^{\frac{3}{2}} + 16 )}{3} + \frac{1792 ( - 2 x ^{\frac{5}{2}} + 64 )}{5} + \frac{16 ( - 2 x ^{\frac{9}{2}} + 1024 )}{9} + 38912

Solution steps

\int_{x}^{4} (2 + u)^{8} d u

Expand

(2 + u)^{8} : 256 + 1024 u + 1792 u + 1792 u^{\frac{3}{2}} + 1120 u^{2} + 448 u^{\frac{5}{2}} + 112 u^{3} + 16 u^{\frac{7}{2}} + u^{4}

= \int_{x}^{4} 256 + 1024 u + 1792 u + 1792 u^{\frac{3}{2}} + 1120 u^{2} + 448 u^{\frac{5}{2}} + 112 u^{3} + 16 u^{\frac{7}{2}} + u^{4} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{x}^{4} 256 d u + \int_{x}^{4} 1024 u d u + \int_{x}^{4} 1792 u d u + \int_{x}^{4} 1792 u^{\frac{3}{2}} d u + \int_{x}^{4} 1120 u^{2} d u + \int_{x}^{4} 448 u^{\frac{5}{2}} d u + \int_{x}^{4} 112 u^{3} d u + \int_{x}^{4} 16 u^{\frac{7}{2}} d u + \int_{x}^{4} u^{4} d u

\int_{x}^{4} 256 d u = 1024 - 256 x

\int_{x}^{4} 1024 u d u = \frac{1024 ( 16 - 2 x ^{\frac{3}{2}} )}{3}

\int_{x}^{4} 1792 u d u = 1792 (8 - \frac{x ^{2}}{2})

\int_{x}^{4} 1792 u^{\frac{3}{2}} d u = \frac{1792 ( 64 - 2 x ^{\frac{5}{2}} )}{5}

\int_{x}^{4} 1120 u^{2} d u = \frac{1120 ( 64 - x ^{3} )}{3}

\int_{x}^{4} 448 u^{\frac{5}{2}} d u = 64 (- 2 x^{\frac{7}{2}} + 256)

\int_{x}^{4} 112 u^{3} d u = 112 (64 - \frac{x ^{4}}{4})

\int_{x}^{4} 16 u^{\frac{7}{2}} d u = \frac{16 ( 1024 - 2 x ^{\frac{9}{2}} )}{9}

\int_{x}^{4} u^{4} d u = \frac{1024 - x ^{5}}{5}

= 1024 - 256 x + \frac{1024 ( 16 - 2 x ^{\frac{3}{2}} )}{3} + 1792 (8 - \frac{x ^{2}}{2}) + \frac{1792 ( 64 - 2 x ^{\frac{5}{2}} )}{5} + \frac{1120 ( 64 - x ^{3} )}{3} + 64 (- 2 x^{\frac{7}{2}} + 256) + 112 (64 - \frac{x ^{4}}{4}) + \frac{16 ( 1024 - 2 x ^{\frac{9}{2}} )}{9} + \frac{1024 - x ^{5}}{5}

Simplify

1024 - 256 x + \frac{1024 ( 16 - 2 x ^{\frac{3}{2}} )}{3} + 1792 (8 - \frac{x ^{2}}{2}) + \frac{1792 ( 64 - 2 x ^{\frac{5}{2}} )}{5} + \frac{1120 ( 64 - x ^{3} )}{3} + 64 (- 2 x^{\frac{7}{2}} + 256) + 112 (64 - \frac{x ^{4}}{4}) + \frac{16 ( 1024 - 2 x ^{\frac{9}{2}} )}{9} + \frac{1024 - x ^{5}}{5} : - 28 x^{4} - 896 x^{2} - 256 x - 128 x^{\frac{7}{2}} + \frac{- x ^{5} + 1024}{5} + \frac{1120 ( - x ^{3} + 64 )}{3} + \frac{1024 ( - 2 x ^{\frac{3}{2}} + 16 )}{3} + \frac{1792 ( - 2 x ^{\frac{5}{2}} + 64 )}{5} + \frac{16 ( - 2 x ^{\frac{9}{2}} + 1024 )}{9} + 38912

= - 28 x^{4} - 896 x^{2} - 256 x - 128 x^{\frac{7}{2}} + \frac{- x ^{5} + 1024}{5} + \frac{1120 ( - x ^{3} + 64 )}{3} + \frac{1024 ( - 2 x ^{\frac{3}{2}} + 16 )}{3} + \frac{1792 ( - 2 x ^{\frac{5}{2}} + 64 )}{5} + \frac{16 ( - 2 x ^{\frac{9}{2}} + 1024 )}{9} + 38912

\int_{- 1}^{1} \frac{1}{∣ x ∣} d x

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What is the integral from x to 4 of (2+sqrt(u))^8 ?
The integral from x to 4 of (2+sqrt(u))^8 is