What is the integral of (4x+1)/(4sqrt(x)) ?

The integral of (4x+1)/(4sqrt(x)) is 1/6 sqrt(x)(4x+3)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (4x+1)/(4sqrt(x))

\int \frac{4 x + 1}{4 x} d x

\frac{1}{6} x (4 x + 3) + C

Solution steps

\int \frac{4 x + 1}{4 x} d x

Take the constant out:

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

= \frac{1}{4} \cdot \int \frac{4 x + 1}{x} d x

Expand

\frac{4 x + 1}{x} : 4 x + \frac{1}{x}

= \frac{1}{4} \cdot \int 4 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

= \frac{1}{4} (\int 4 x d x + \int \frac{1}{x} d x)

\int 4 x d x = \frac{8}{3} x^{\frac{3}{2}}

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

= \frac{1}{4} (\frac{8}{3} x^{\frac{3}{2}} + 2 x)

Simplify

\frac{1}{4} (\frac{8}{3} x^{\frac{3}{2}} + 2 x) : \frac{1}{6} x (4 x + 3)

= \frac{1}{6} x (4 x + 3)

Add a constant to the solution

= \frac{1}{6} x (4 x + 3) + C

\int \frac{d u}{u ^{2}} d x

\int 3 e^{x} + π^{2} d x

\int - \frac{8}{x ^{6}} d x

\int \frac{2 x}{x ^{4} - 4} d x

\int \frac{x}{( r ^{2} + x ^{2} ) ^{\frac{3}{2}}} d x

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