What is the integral of (9x^5-6x^3+4x) ?

The integral of (9x^5-6x^3+4x) is (3x^6)/2-(3x^4)/2+2x^2+C

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integral of (9x^5-6x^3+4x)

\int (9 x^{5} - 6 x^{3} + 4 x) d x

\frac{3 x ^{6}}{2} - \frac{3 x ^{4}}{2} + 2 x^{2} + C

Solution steps

\int 9 x^{5} - 6 x^{3} + 4 x 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 9 x^{5} d x - \int 6 x^{3} d x + \int 4 x d x

\int 9 x^{5} d x = \frac{3 x ^{6}}{2}

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

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

= \frac{3 x ^{6}}{2} - \frac{3 x ^{4}}{2} + 2 x^{2}

Add a constant to the solution

= \frac{3 x ^{6}}{2} - \frac{3 x ^{4}}{2} + 2 x^{2} + C

10 x

\frac{d}{d x} (1 + cot (x))

x \to 1 lim (\frac{1 - \frac{1}{x}}{x - 1})

\frac{d}{d x} (\frac{cos ( x - y )}{sin ( y - x )})

\frac{d}{d x} (2 x^{4} - 8 x + 3)

What is the integral of (9x^5-6x^3+4x) ?
The integral of (9x^5-6x^3+4x) is (3x^6)/2-(3x^4)/2+2x^2+C