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

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

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

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

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

Solution steps

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

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

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

Expand

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

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

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

Add a constant to the solution

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

\int x^{3} 81 - x^{2} d x

\frac{\partial}{\partial x} (x + ln (y))

y^{^{'}} - y = 6 t e^{2 t}, y (0) = 1

x \to 0 lim (x^{2} \cdot e^{- x})

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

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