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

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

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

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

\frac{1}{2} (\frac{x ^{5}}{5} - \frac{1}{3 x ^{3}}) + C

Solution steps

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

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{x ^{8} + 1}{x ^{4}} d x

Expand

\frac{x ^{8} + 1}{x ^{4}} : x^{4} + \frac{1}{x ^{4}}

= \frac{1}{2} \cdot \int x^{4} + \frac{1}{x ^{4}} 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

= \frac{1}{2} (\int x^{4} d x + \int \frac{1}{x ^{4}} d x)

\int x^{4} d x = \frac{x ^{5}}{5}

\int \frac{1}{x ^{4}} d x = - \frac{1}{3 x ^{3}}

= \frac{1}{2} (\frac{x ^{5}}{5} - \frac{1}{3 x ^{3}})

Add a constant to the solution

= \frac{1}{2} (\frac{x ^{5}}{5} - \frac{1}{3 x ^{3}}) + C

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