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

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

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

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

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

Solution steps

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

Multiply by the conjugate of

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

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

Simplify

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

Apply the Sum Rule:

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

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

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

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

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

Simplify

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

Add a constant to the solution

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

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