What is the integral of ((18x^6+3x^4)/(6x^2)) ?

The integral of ((18x^6+3x^4)/(6x^2)) is 1/6 ((18x^5)/5+x^3)+C

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integral of ((18x^6+3x^4)/(6x^2))

\int (\frac{18 x ^{6} + 3 x ^{4}}{6 x ^{2}}) d x

\frac{1}{6} (\frac{18 x ^{5}}{5} + x^{3}) + C

Solution steps

\int \frac{18 x ^{6} + 3 x ^{4}}{6 x ^{2}} d x

Take the constant out:

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

= \frac{1}{6} \cdot \int \frac{18 x ^{6} + 3 x ^{4}}{x ^{2}} d x

Expand

\frac{18 x ^{6} + 3 x ^{4}}{x ^{2}} : \frac{18 x ^{6}}{x ^{2}} + \frac{3 x ^{4}}{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}{6} (\int \frac{18 x ^{6}}{x ^{2}} d x + \int \frac{3 x ^{4}}{x ^{2}} d x)

\int \frac{18 x ^{6}}{x ^{2}} d x = \frac{18 x ^{5}}{5}

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

= \frac{1}{6} (\frac{18 x ^{5}}{5} + x^{3})

Add a constant to the solution

= \frac{1}{6} (\frac{18 x ^{5}}{5} + x^{3}) + C

\int_{0}^{\frac{π}{2}} 42

y = x^{3} + 4 x - 3, (1, 2)

4 e^{a r c c o s (4 x)}

\int 3 x^{3} d x

What is the integral of ((18x^6+3x^4)/(6x^2)) ?
The integral of ((18x^6+3x^4)/(6x^2)) is 1/6 ((18x^5)/5+x^3)+C