What is the integral of 8x+1/(sqrt(x)) ?

The integral of 8x+1/(sqrt(x)) is 4x^2+2sqrt(x)+C

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integral of 8x+1/(sqrt(x))

\int 8 x + \frac{1}{x} d x

4 x^{2} + 2 x + C

Solution steps

\int 8 x + \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 8 x d x + \int \frac{1}{x} d x

\int 8 x d x = 4 x^{2}

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

= 4 x^{2} + 2 x

Add a constant to the solution

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

\frac{d}{d x} (\frac{x + 4}{3 x ^{2} e ^{x}})

\int_{0}^{1} e^{x} d x

\frac{d}{d x} (x + a)

\frac{d}{d x} (ln (x - y))

What is the integral of 8x+1/(sqrt(x)) ?
The integral of 8x+1/(sqrt(x)) is 4x^2+2sqrt(x)+C