What is the integral of (sqrt(x)+1/(2sqrt(x)))^2 ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of (sqrt(x)+1/(2sqrt(x)))^2

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

\frac{x ^{2}}{2} + x + \frac{1}{4} ln ∣ x ∣ + C

Solution steps

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

Expand

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

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

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

\int 1 d x = x

\int \frac{1}{4 x} d x = \frac{1}{4} ln ∣ x ∣

= \frac{x ^{2}}{2} + x + \frac{1}{4} ln ∣ x ∣

Add a constant to the solution

= \frac{x ^{2}}{2} + x + \frac{1}{4} ln ∣ x ∣ + C

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