What is the integral of ((x^5+1)^2)/(2x^2) ?

The integral of ((x^5+1)^2)/(2x^2) is 1/2 ((x^9)/9+(x^4)/2-1/x)+C

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integral of ((x^5+1)^2)/(2x^2)

\int \frac{( x ^{5} + 1 ) ^{2}}{2 x ^{2}} d x

\frac{1}{2} (\frac{x ^{9}}{9} + \frac{x ^{4}}{2} - \frac{1}{x}) + C

Solution steps

\int \frac{( x ^{5} + 1 ) ^{2}}{2 x ^{2}} 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 ^{5} + 1 ) ^{2}}{x ^{2}} d x

Expand

\frac{( x ^{5} + 1 ) ^{2}}{x ^{2}} : x^{8} + 2 x^{3} + \frac{1}{x ^{2}}

= \frac{1}{2} \cdot \int x^{8} + 2 x^{3} + \frac{1}{x ^{2}} 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^{8} d x + \int 2 x^{3} d x + \int \frac{1}{x ^{2}} d x)

\int x^{8} d x = \frac{x ^{9}}{9}

\int 2 x^{3} d x = \frac{x ^{4}}{2}

\int \frac{1}{x ^{2}} d x = - \frac{1}{x}

= \frac{1}{2} (\frac{x ^{9}}{9} + \frac{x ^{4}}{2} - \frac{1}{x})

Add a constant to the solution

= \frac{1}{2} (\frac{x ^{9}}{9} + \frac{x ^{4}}{2} - \frac{1}{x}) + C

What is the integral of ((x^5+1)^2)/(2x^2) ?
The integral of ((x^5+1)^2)/(2x^2) is 1/2 ((x^9)/9+(x^4)/2-1/x)+C