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

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

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

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

2 ln ∣ x ∣ + C

Solution steps

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

Simplify

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

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

Take the constant out:

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

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

Simplify

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

x^{2} = (x),

assuming

x \geq 0

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

Use the common integral:

\int \frac{1}{x} d x = ln (∣ x ∣)

= 2 ln ∣ x ∣

Add a constant to the solution

= 2 ln ∣ x ∣ + C

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