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

The integral of (6x^5-4x^3+1/(sqrt(x))) is x^6-x^4+2sqrt(x)+C

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

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

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

Solution steps

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

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

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

\int \frac{1}{x} d x = 2 x

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

Add a constant to the solution

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

4 e (cos (x) - x sin (x))

x \to 4 lim (\frac{6}{x - 4})

x \to 3 lim (2 x + ∣ x + 3 ∣)

\int \frac{8}{9} x^{\frac{8}{9}} d x

\frac{d}{d y} (\frac{e ^{x}}{y})

What is the integral of (6x^5-4x^3+1/(sqrt(x))) ?
The integral of (6x^5-4x^3+1/(sqrt(x))) is x^6-x^4+2sqrt(x)+C