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

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

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integral of (x^4-6x^2+3)

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

\frac{x ^{5}}{5} - 2 x^{3} + 3 x + C

Solution steps

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

\int x^{4} d x = \frac{x ^{5}}{5}

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

\int 3 d x = 3 x

= \frac{x ^{5}}{5} - 2 x^{3} + 3 x

Add a constant to the solution

= \frac{x ^{5}}{5} - 2 x^{3} + 3 x + C

\int \frac{3 x ^{4} - 12}{x ^{2} + 2} d x

\int x^{2} - 2 x - 3 d x

\int arcsin (\frac{x}{1 + x}) d x

\int tan (\frac{3 x}{7}) d x

\int \frac{x}{1 - cos ( x )} d x

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