What is the integral of ((6x^2+2sqrt(x))/(2sqrt(x))) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of ((6x^2+2sqrt(x))/(2sqrt(x)))

\int (\frac{6 x ^{2} + 2 x}{2 x}) d x

\frac{1}{5} (6 x^{\frac{5}{2}} + 5 x) + C

Solution steps

\int \frac{6 x ^{2} + 2 x}{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 ^{2} + 2 x}{x} d x

Expand

\frac{6 x ^{2} + 2 x}{x} : 6 x^{\frac{3}{2}} + 2

= \frac{1}{2} \cdot \int 6 x^{\frac{3}{2}} + 2 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^{\frac{3}{2}} d x + \int 2 d x)

\int 6 x^{\frac{3}{2}} d x = \frac{12}{5} x^{\frac{5}{2}}

\int 2 d x = 2 x

= \frac{1}{2} (\frac{12}{5} x^{\frac{5}{2}} + 2 x)

Simplify

\frac{1}{2} (\frac{12}{5} x^{\frac{5}{2}} + 2 x) : \frac{1}{5} (6 x^{\frac{5}{2}} + 5 x)

= \frac{1}{5} (6 x^{\frac{5}{2}} + 5 x)

Add a constant to the solution

= \frac{1}{5} (6 x^{\frac{5}{2}} + 5 x) + C

\int 3^{2 x + 4} + ln (3 - x) - e^{2 - 3 x} d x

\int sin (2 x) e^{- 3 x} d x

\int x 8 - 4 x d x

\int x 36 - x d x

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

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