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

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

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

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

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

Solution steps

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

x^{2} = (x),

assuming

x \geq 0

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

x^{2} x = x^{3}

= \int \frac{1}{x ^{3}} d x

Apply the Power Rule

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

= - \frac{1}{2 x ^{2}}

Add a constant to the solution

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

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