What is the integral of (2+x^8)^48x^7 ?

The integral of (2+x^8)^48x^7 is 8(2x^8+2x^{16}+x^{24}+(x^{32})/4+(x^{40})/(40))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (2+x^8)^48x^7

\int (2 + x^{8})^{4} 8 x^{7} d x

8 (2 x^{8} + 2 x^{16} + x^{24} + \frac{x ^{32}}{4} + \frac{x ^{40}}{40}) + C

Solution steps

\int (2 + x^{8})^{4} \cdot 8 x^{7} d x

Take the constant out:

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

= 8 \cdot \int (2 + x^{8})^{4} x^{7} d x

Expand

(2 + x^{8})^{4} x^{7} : 16 x^{7} + 32 x^{15} + 24 x^{23} + 8 x^{31} + x^{39}

= 8 \cdot \int 16 x^{7} + 32 x^{15} + 24 x^{23} + 8 x^{31} + x^{39} 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

= 8 (\int 16 x^{7} d x + \int 32 x^{15} d x + \int 24 x^{23} d x + \int 8 x^{31} d x + \int x^{39} d x)

\int 16 x^{7} d x = 2 x^{8}

\int 32 x^{15} d x = 2 x^{16}

\int 24 x^{23} d x = x^{24}

\int 8 x^{31} d x = \frac{x ^{32}}{4}

\int x^{39} d x = \frac{x ^{40}}{40}

= 8 (2 x^{8} + 2 x^{16} + x^{24} + \frac{x ^{32}}{4} + \frac{x ^{40}}{40})

Add a constant to the solution

= 8 (2 x^{8} + 2 x^{16} + x^{24} + \frac{x ^{32}}{4} + \frac{x ^{40}}{40}) + C

\int \frac{1}{x ^{2} - 6 x + 13} d x

f (x) = 70 (4 + 0.1 x + 0.01 x^{2})

\int \frac{18}{sec ( 2 x ) - 1} d x

\int \frac{1}{e ^{x} + e ^{- x} + 2} d x

8 y^{^{''}} + y^{^{'}} = 0

What is the integral of (2+x^8)^48x^7 ?
The integral of (2+x^8)^48x^7 is 8(2x^8+2x^{16}+x^{24}+(x^{32})/4+(x^{40})/(40))+C