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

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

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

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

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

Solution steps

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

Apply Integration By Parts

: \frac{x}{( x ^{2} - 1 ) ^{\frac{1}{2}}} - \int - \frac{x ^{2}}{( x ^{2} - 1 ) ^{\frac{3}{2}}} d x

= \frac{x}{( x ^{2} - 1 ) ^{\frac{1}{2}}} - \int - \frac{x ^{2}}{( x ^{2} - 1 ) ^{\frac{3}{2}}} d x

\int - \frac{x ^{2}}{( x ^{2} - 1 ) ^{\frac{3}{2}}} d x = \frac{x}{( x ^{2} - 1 ) ^{\frac{1}{2}}} - ln (x^{2} - 1)^{\frac{1}{2}} + x

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

Simplify

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

Add a constant to the solution

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

\frac{d y}{d x} = \frac{3}{x y ^{2}}

\int \frac{cos ( 2 ln ( x ) ) ln ( x )}{x} d x

\int (x^{3} + x^{2}) d x

f (x) = 7 + 5 x

\frac{\partial}{\partial y} (x^{2} y - x y^{3})

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