What is the integral of (18x^5-12x^3+4x) ?

The integral of (18x^5-12x^3+4x) is 3x^6-3x^4+2x^2+C

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integral of (18x^5-12x^3+4x)

\int (18 x^{5} - 12 x^{3} + 4 x) d x

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

Solution steps

\int 18 x^{5} - 12 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 18 x^{5} d x - \int 12 x^{3} d x + \int 4 x d x

\int 18 x^{5} d x = 3 x^{6}

\int 12 x^{3} d x = 3 x^{4}

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

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

Add a constant to the solution

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

\int x^{5} 1 - x^{3} d x

\int \frac{y ^{3} - 4 y + 3}{y + 5} d y

\int \frac{1}{x + 2 x ^{3}} d x

\int \frac{1}{2 x ^{2} x ^{2} - 16} d x

\int \frac{7}{w ^{2} + 3 w + 3} d w

What is the integral of (18x^5-12x^3+4x) ?
The integral of (18x^5-12x^3+4x) is 3x^6-3x^4+2x^2+C