What is the integral of sqrt((1+1)/(x^4)) ?

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

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integral of sqrt((1+1)/(x^4))

integral

\frac{1 + 1}{x ^{4}}

- \frac{2}{x} + C

Solution steps

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

Simplify

\frac{1 + 1}{x ^{4}} : \frac{1 + 1}{x ^{4}}

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

Simplify

= 2 \cdot \int \frac{1}{x ^{4}} d x

x^{4} = x^{2},

assuming

x \geq 0

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

Apply the Power Rule

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

Simplify

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

= - \frac{2}{x}

Add a constant to the solution

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

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