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

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

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integral of (3x^4-5x^2+6x)

\int (3 x^{4} - 5 x^{2} + 6 x) d x

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

Solution steps

\int 3 x^{4} - 5 x^{2} + 6 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 3 x^{4} d x - \int 5 x^{2} d x + \int 6 x d x

\int 3 x^{4} d x = \frac{3 x ^{5}}{5}

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

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

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

Add a constant to the solution

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

x \to 0 lim (\frac{∣ x + 1 ∣ - ∣ x - 1 ∣}{x})

n = 1 \sum \infty 8

\frac{\partial}{\partial v} (u^{2} + v^{2})

1 - e^{- 0.4 t}

\int \frac{1}{5 + 2 x} d x

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