What is the integral from 1 to 2 of (x^5+5x^8)/(x^4) ?

The integral from 1 to 2 of (x^5+5x^8)/(x^4) is 65/2

integral from 1 to 2 of (x^5+5x^8)/(x^4)

\int_{1}^{2} \frac{x ^{5} + 5 x ^{8}}{x ^{4}} d x

\frac{65}{2}

Decimal

32.5

Solution steps

\int_{1}^{2} \frac{x ^{5} + 5 x ^{8}}{x ^{4}} d x

Expand

\frac{x ^{5} + 5 x ^{8}}{x ^{4}} : \frac{x ^{5}}{x ^{4}} + \frac{5 x ^{8}}{x ^{4}}

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}^{2} \frac{x ^{5}}{x ^{4}} d x + \int_{1}^{2} \frac{5 x ^{8}}{x ^{4}} d x

\int_{1}^{2} \frac{x ^{5}}{x ^{4}} d x = \frac{3}{2}

\int_{1}^{2} \frac{5 x ^{8}}{x ^{4}} d x = 31

= \frac{3}{2} + 31

Simplify

\frac{3}{2} + 31 : \frac{65}{2}

= \frac{65}{2}

\frac{d}{d x} (x^{\frac{1}{3}} + 2 x)

\int_{0}^{9} x d x

y^{^{''}} + 4 y = t - 2 sin (2 t)

\frac{d}{d x} (\frac{( x - 1 )}{x + 1})

x \to \infty lim (\frac{x ^{2} - 9}{2 x - 6})

What is the integral from 1 to 2 of (x^5+5x^8)/(x^4) ?
The integral from 1 to 2 of (x^5+5x^8)/(x^4) is 65/2