What is the integral of (6x-5sqrt(x)+1)/(2sqrt(x)) ?

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

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integral of (6x-5sqrt(x)+1)/(2sqrt(x))

\int \frac{6 x - 5 x + 1}{2 x} d x

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

Solution steps

\int \frac{6 x - 5 x + 1}{2 x} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int \frac{6 x - 5 x + 1}{x} d x

Expand

\frac{6 x - 5 x + 1}{x} : 6 x - 5 + \frac{1}{x}

= \frac{1}{2} \cdot \int 6 x - 5 + \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

= \frac{1}{2} (\int 6 x d x - \int 5 d x + \int \frac{1}{x} d x)

\int 6 x d x = 4 x^{\frac{3}{2}}

\int 5 d x = 5 x

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

= \frac{1}{2} (4 x^{\frac{3}{2}} - 5 x + 2 x)

Simplify

\frac{1}{2} (4 x^{\frac{3}{2}} - 5 x + 2 x) : 2 x^{\frac{3}{2}} - \frac{5}{2} x + x

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

Add a constant to the solution

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

\frac{d}{d x} ((4 x + 5)^{3})

\int_{0}^{1} (\frac{x - x}{3}) d x

cot (x^{6})

\frac{1}{y} \frac{d y}{d x} = e^{x} + lo g_{e} (y)

\frac{d}{dd} (d cos (t) + t^{2} sin (t))

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