What is the integral from 1 to 2 of (x^5+4x^7)/(x^4) ?

The integral from 1 to 2 of (x^5+4x^7)/(x^4) is 33/2

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integral from 1 to 2 of (x^5+4x^7)/(x^4)

\int_{1}^{2} \frac{x ^{5} + 4 x ^{7}}{x ^{4}} d x

\frac{33}{2}

Solution steps

\int_{1}^{2} \frac{x ^{5} + 4 x ^{7}}{x ^{4}} d x

Expand

\frac{x ^{5} + 4 x ^{7}}{x ^{4}} : \frac{x ^{5}}{x ^{4}} + \frac{4 x ^{7}}{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{4 x ^{7}}{x ^{4}} d x

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

\int_{1}^{2} \frac{4 x ^{7}}{x ^{4}} d x = 15

= \frac{3}{2} + 15

Simplify

\frac{3}{2} + 15 : \frac{33}{2}

= \frac{33}{2}

\frac{\partial}{\partial x} (cos (x yz))

(x^{5} + 2)^{2} (x^{4} + 4)^{4}

\int_{6}^{7} \frac{x}{x ^{2} - 1} d x

(1 + ln (x) + \frac{y}{x}) d x = (2 - ln (x)) d y

\frac{\partial}{\partial x} (e^{- 3 x - y})

What is the integral from 1 to 2 of (x^5+4x^7)/(x^4) ?
The integral from 1 to 2 of (x^5+4x^7)/(x^4) is 33/2