What is the integral of ((6x^512x^4+9x^3)/(3x^2)) ?

The integral of ((6x^512x^4+9x^3)/(3x^2)) is 3/2 x^2(2x^6+1)+C

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integral of ((6x^512x^4+9x^3)/(3x^2))

\int (\frac{6 x ^{5} 12 x ^{4} + 9 x ^{3}}{3 x ^{2}}) d x

\frac{3}{2} x^{2} (2 x^{6} + 1) + C

Solution steps

\int \frac{6 x ^{5} \cdot 12 x ^{4} + 9 x ^{3}}{3 x ^{2}} d x

Take the constant out:

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

= \frac{1}{3} \cdot \int \frac{6 x ^{5} \cdot 12 x ^{4} + 9 x ^{3}}{x ^{2}} d x

Expand

\frac{6 x ^{5} 12 x ^{4} + 9 x ^{3}}{x ^{2}} : \frac{6 x ^{5} \cdot 12 x ^{4}}{x ^{2}} + \frac{9 x ^{3}}{x ^{2}}

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}{3} (\int \frac{6 x ^{5} \cdot 12 x ^{4}}{x ^{2}} d x + \int \frac{9 x ^{3}}{x ^{2}} d x)

\int \frac{6 x ^{5} \cdot 12 x ^{4}}{x ^{2}} d x = 9 x^{8}

\int \frac{9 x ^{3}}{x ^{2}} d x = \frac{9 x ^{2}}{2}

= \frac{1}{3} (9 x^{8} + \frac{9 x ^{2}}{2})

Simplify

\frac{1}{3} (9 x^{8} + \frac{9 x ^{2}}{2}) : \frac{3}{2} x^{2} (2 x^{6} + 1)

= \frac{3}{2} x^{2} (2 x^{6} + 1)

Add a constant to the solution

= \frac{3}{2} x^{2} (2 x^{6} + 1) + C

What is the integral of ((6x^512x^4+9x^3)/(3x^2)) ?
The integral of ((6x^512x^4+9x^3)/(3x^2)) is 3/2 x^2(2x^6+1)+C