What is the integral of 1/(x^2)*(1/t-1)^{1/2} ?

The integral of 1/(x^2)*(1/t-1)^{1/2} is -((-t+1)^{1/2})/(t^{1/2)x}+C

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integral of 1/(x^2)*(1/t-1)^{1/2}

\int \frac{1}{x ^{2}} \cdot (\frac{1}{t} - 1)^{\frac{1}{2}}

- \frac{( - t + 1 ) ^{\frac{1}{2}}}{t ^{\frac{1}{2}} x} + C

Solution steps

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

Take the constant out:

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

= (- 1 + \frac{1}{t})^{\frac{1}{2}} \cdot \int \frac{1}{x ^{2}} d x

Apply the Power Rule

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

Simplify

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

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

Add a constant to the solution

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

What is the integral of 1/(x^2)*(1/t-1)^{1/2} ?
The integral of 1/(x^2)*(1/t-1)^{1/2} is -((-t+1)^{1/2})/(t^{1/2)x}+C