What is the integral of (1/4 x^6-5x^3+9x) ?

The integral of (1/4 x^6-5x^3+9x) is (x^7}{28}-\frac{5x^4)/4+(9x^2)/2+C

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integral of (1/4 x^6-5x^3+9x)

\int (\frac{1}{4} x^{6} - 5 x^{3} + 9 x) d x

\frac{x ^{7}}{28} - \frac{5 x ^{4}}{4} + \frac{9 x ^{2}}{2} + C

Solution steps

\int \frac{1}{4} x^{6} - 5 x^{3} + 9 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 \frac{1}{4} x^{6} d x - \int 5 x^{3} d x + \int 9 x d x

\int \frac{1}{4} x^{6} d x = \frac{x ^{7}}{28}

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

\int 9 x d x = \frac{9 x ^{2}}{2}

= \frac{x ^{7}}{28} - \frac{5 x ^{4}}{4} + \frac{9 x ^{2}}{2}

Add a constant to the solution

= \frac{x ^{7}}{28} - \frac{5 x ^{4}}{4} + \frac{9 x ^{2}}{2} + C

\frac{\partial}{\partial x} (- (xx - yy) e^{- xx - yy})

x \to \infty lim (\frac{e ^{x^{3}}}{7 - x ^{4}})

f^{^{'}} (x) = \frac{1}{x}

y = x^{3} - 2 x + 1

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

What is the integral of (1/4 x^6-5x^3+9x) ?
The integral of (1/4 x^6-5x^3+9x) is (x^7}{28}-\frac{5x^4)/4+(9x^2)/2+C