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

The integral of (2x)/(x+sqrt(x^2-1)) is 2/3 x^3-x^2sqrt(x^2-1)+1/3 (x^2-1)^{3/2}+sqrt(x^2-1)+C

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

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

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

Solution steps

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

Take the constant out:

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

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

Apply Integration By Parts

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

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

\int \frac{1}{2} (x^{2} - x x^{2} - 1 + ln x + x^{2} - 1) d x = \frac{1}{6} (x^{3} - (x^{2} - 1)^{\frac{3}{2}} + 3 x ln x + x^{2} - 1 - 3 x^{2} - 1)

= 2 (x (\frac{x ^{2}}{2} - \frac{1}{2} x x^{2} - 1 + \frac{1}{2} ln x + x^{2} - 1) - \frac{1}{6} (x^{3} - (x^{2} - 1)^{\frac{3}{2}} + 3 x ln x + x^{2} - 1 - 3 x^{2} - 1))

Simplify

2 (x (\frac{x ^{2}}{2} - \frac{1}{2} x x^{2} - 1 + \frac{1}{2} ln x + x^{2} - 1) - \frac{1}{6} (x^{3} - (x^{2} - 1)^{\frac{3}{2}} + 3 x ln x + x^{2} - 1 - 3 x^{2} - 1)) : \frac{2}{3} x^{3} - x^{2} x^{2} - 1 + \frac{1}{3} (x^{2} - 1)^{\frac{3}{2}} + x^{2} - 1

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

Add a constant to the solution

= \frac{2}{3} x^{3} - x^{2} x^{2} - 1 + \frac{1}{3} (x^{2} - 1)^{\frac{3}{2}} + x^{2} - 1 + C

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What is the integral of (2x)/(x+sqrt(x^2-1)) ?
The integral of (2x)/(x+sqrt(x^2-1)) is 2/3 x^3-x^2sqrt(x^2-1)+1/3 (x^2-1)^{3/2}+sqrt(x^2-1)+C