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

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

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

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

\frac{x ^{2}}{2} + 2 x + ln ∣ x ∣ + C

Solution steps

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

Expand

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

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

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

\int 2 d x = 2 x

\int \frac{1}{x} d x = ln ∣ x ∣

= \frac{x ^{2}}{2} + 2 x + ln ∣ x ∣

Add a constant to the solution

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

\frac{\partial}{\partial z} (e^{z^{2}})

\int 7 \frac{cos ( x ) + sin ( x )}{sin ( 2 x )} d x

\frac{\partial}{\partial x} (x^{y^{2}})

\frac{\partial}{\partial y} (3 x^{2} y^{2} - 2)

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

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