What is the integral of (2x+3sqrt(x))/(4x^2) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (2x+3sqrt(x))/(4x^2)

\int \frac{2 x + 3 x}{4 x ^{2}} d x

\frac{1}{4} (2 ln ∣ x ∣ - \frac{6}{x}) + C

Solution steps

\int \frac{2 x + 3 x}{4 x ^{2}} 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{2 x + 3 x}{x ^{2}} d x

Simplify

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

= \frac{1}{4} \cdot \int \frac{2 x + 3}{x ^{\frac{3}{2}}} d x

Expand

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

= \frac{1}{4} \cdot \int \frac{2}{x} + \frac{3}{x ^{\frac{3}{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}{4} (\int \frac{2}{x} d x + \int \frac{3}{x ^{\frac{3}{2}}} d x)

\int \frac{2}{x} d x = 2 ln ∣ x ∣

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

= \frac{1}{4} (2 ln ∣ x ∣ - \frac{6}{x})

Add a constant to the solution

= \frac{1}{4} (2 ln ∣ x ∣ - \frac{6}{x}) + C

\int cos^{6} (y) sin^{3} (y) d y

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

\int \frac{1}{x ^{2} ( a ^{2} - x ^{2} ) ^{\frac{1}{2}}} d x

\int e^{- x} sin (π x) d x

\int \frac{x - 3}{x ^{2} + 3 x + 3} d x

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