What is the integral of (42w^7-7w)/(7w^5) ?

The integral of (42w^7-7w)/(7w^5) is (6w^6+1)/(3w^3)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (42w^7-7w)/(7w^5)

\int \frac{42 w ^{7} - 7 w}{7 w ^{5}} d w

\frac{6 w ^{6} + 1}{3 w ^{3}} + C

Solution steps

\int \frac{42 w ^{7} - 7 w}{7 w ^{5}} d w

Take the constant out:

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

= \frac{1}{7} \cdot \int \frac{42 w ^{7} - 7 w}{w ^{5}} d w

Expand

\frac{42 w ^{7} - 7 w}{w ^{5}} : 42 w^{2} - \frac{7}{w ^{4}}

= \frac{1}{7} \cdot \int 42 w^{2} - \frac{7}{w ^{4}} d w

Apply the Sum Rule:

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

= \frac{1}{7} (\int 42 w^{2} d w - \int \frac{7}{w ^{4}} d w)

\int 42 w^{2} d w = 14 w^{3}

\int \frac{7}{w ^{4}} d w = - \frac{7}{3 w ^{3}}

= \frac{1}{7} (14 w^{3} - (- \frac{7}{3 w ^{3}}))

Simplify

\frac{1}{7} (14 w^{3} - (- \frac{7}{3 w ^{3}})) : \frac{6 w ^{6} + 1}{3 w ^{3}}

= \frac{6 w ^{6} + 1}{3 w ^{3}}

Add a constant to the solution

= \frac{6 w ^{6} + 1}{3 w ^{3}} + C

x \to 2 lim (\frac{x ^{3} - 8}{x + 2})

(4, 0), (- 2, 4)

y^{^{''}} + 4 y^{^{'}} + 3 y = 4 e^{- t}, y (0) = 0, y^{^{'}} (0) = 2

\frac{d}{d x} (\frac{1}{sin ( x )})

f (x) = ln (5)

What is the integral of (42w^7-7w)/(7w^5) ?
The integral of (42w^7-7w)/(7w^5) is (6w^6+1)/(3w^3)+C