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

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

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integral of ((3x^4-3x^2+x))/x

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

\frac{3 x ^{4}}{4} - \frac{3 x ^{2}}{2} + x + C

Solution steps

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

Simplify

\frac{3 x ^{4} - 3 x ^{2} + x}{x} : 3 x^{3} - 3 x + 1

= \int 3 x^{3} - 3 x + 1 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

= \int 3 x^{3} d x - \int 3 x d x + \int 1 d x

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

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

\int 1 d x = x

= \frac{3 x ^{4}}{4} - \frac{3 x ^{2}}{2} + x

Add a constant to the solution

= \frac{3 x ^{4}}{4} - \frac{3 x ^{2}}{2} + x + C

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