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

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

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integral of sqrt(1+2/x)

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

x 2 + x + ln \frac{2 + x}{x} + 1 - ln \frac{2 + x}{x} - 1 + C

Solution steps

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

Apply u-substitution

= \int - \frac{2 u + 1}{u ^{2}} d u

Take the constant out:

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

= - 2 \cdot \int \frac{u + 1}{u ^{2}} d u

Apply Integration By Parts

= - 2 (- \frac{u + 1}{u} - \int - \frac{1}{2 u u + 1} d u)

\int - \frac{1}{2 u u + 1} d u = \frac{1}{2} ln u + 1 + 1 - \frac{1}{2} ln u + 1 - 1

= - 2 (- \frac{u + 1}{u} - (\frac{1}{2} ln u + 1 + 1 - \frac{1}{2} ln u + 1 - 1))

Substitute back

u = \frac{2}{x}

= - 2 - \frac{\frac{2}{x} + 1}{\frac{2}{x}} - (\frac{1}{2} ln \frac{2}{x} + 1 + 1 - \frac{1}{2} ln \frac{2}{x} + 1 - 1)

Simplify

- 2 - \frac{\frac{2}{x} + 1}{\frac{2}{x}} - (\frac{1}{2} ln \frac{2}{x} + 1 + 1 - \frac{1}{2} ln \frac{2}{x} + 1 - 1) : x 2 + x + ln \frac{2 + x}{x} + 1 - ln \frac{2 + x}{x} - 1

= x 2 + x + ln \frac{2 + x}{x} + 1 - ln \frac{2 + x}{x} - 1

Add a constant to the solution

= x 2 + x + ln \frac{2 + x}{x} + 1 - ln \frac{2 + x}{x} - 1 + C

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