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

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

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

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

2 x + 42 x + ln ∣ x ∣ + C

Solution steps

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

Simplify

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

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

Expand

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

= \int 2 + \frac{2 2}{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

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

\int 2 d x = 2 x

\int \frac{2 2}{x} d x = 42 x

\int \frac{1}{x} d x = ln ∣ x ∣

= 2 x + 42 x + ln ∣ x ∣

Add a constant to the solution

= 2 x + 42 x + ln ∣ x ∣ + C

n = 1 \sum \infty (\frac{2}{5})^{n}

\frac{sin ( x ^{2} )}{3 x}

\frac{1 + x}{1 - x}

n = 0 \sum \infty \frac{1}{e ^{2 n}}

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

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