What is the integral of (9w^8-5w^5+3w^2-16)/(8w^4) ?

The integral of (9w^8-5w^5+3w^2-16)/(8w^4) is 1/8 ((9w^5)/5-(5w^2)/2-3/w+(16)/(3w^3))+C

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integral of (9w^8-5w^5+3w^2-16)/(8w^4)

\int \frac{9 w ^{8} - 5 w ^{5} + 3 w ^{2} - 16}{8 w ^{4}} d w

\frac{1}{8} (\frac{9 w ^{5}}{5} - \frac{5 w ^{2}}{2} - \frac{3}{w} + \frac{16}{3 w ^{3}}) + C

Solution steps

\int \frac{9 w ^{8} - 5 w ^{5} + 3 w ^{2} - 16}{8 w ^{4}} d w

Take the constant out:

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

= \frac{1}{8} \cdot \int \frac{9 w ^{8} - 5 w ^{5} + 3 w ^{2} - 16}{w ^{4}} d w

Expand

\frac{9 w ^{8} - 5 w ^{5} + 3 w ^{2} - 16}{w ^{4}} : 9 w^{4} - 5 w + \frac{3}{w ^{2}} - \frac{16}{w ^{4}}

= \frac{1}{8} \cdot \int 9 w^{4} - 5 w + \frac{3}{w ^{2}} - \frac{16}{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}{8} (\int 9 w^{4} d w - \int 5 w d w + \int \frac{3}{w ^{2}} d w - \int \frac{16}{w ^{4}} d w)

\int 9 w^{4} d w = \frac{9 w ^{5}}{5}

\int 5 w d w = \frac{5 w ^{2}}{2}

\int \frac{3}{w ^{2}} d w = - \frac{3}{w}

\int \frac{16}{w ^{4}} d w = - \frac{16}{3 w ^{3}}

= \frac{1}{8} (\frac{9 w ^{5}}{5} - \frac{5 w ^{2}}{2} - \frac{3}{w} - (- \frac{16}{3 w ^{3}}))

Simplify

= \frac{1}{8} (\frac{9 w ^{5}}{5} - \frac{5 w ^{2}}{2} - \frac{3}{w} + \frac{16}{3 w ^{3}})

Add a constant to the solution

= \frac{1}{8} (\frac{9 w ^{5}}{5} - \frac{5 w ^{2}}{2} - \frac{3}{w} + \frac{16}{3 w ^{3}}) + C

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What is the integral of (9w^8-5w^5+3w^2-16)/(8w^4) ?
The integral of (9w^8-5w^5+3w^2-16)/(8w^4) is 1/8 ((9w^5)/5-(5w^2)/2-3/w+(16)/(3w^3))+C