How do you find the integral $ln (x + \sqrt{x^{2} - 1})$ $ln( x + sqrt(x^2 -1))$ ?

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How do you find the integral $ln (x + \sqrt{x^{2} - 1})$ $ln( x + sqrt(x^2 -1))$ ?

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Answer 1 · 2018-03-14T01:19:28

$\int ln (x + \sqrt{x^{2} - 1}) d x = x ln (x + \sqrt{x^{2} - 1}) - \sqrt{x^{2} - 1} + C$ $int ln(x+sqrt(x^2-1)) dx = x ln(x+sqrt(x^2-1)) - sqrt(x^2-1) + C$

Explanation:

Use a hyperbolic substitution.

Put:

$x = cosh u$ $x = cosh u$

Then:

$\frac{d x}{d u} = sinh u$ $dx/(du) = sinh u$

and:

$u = ln (x + \sqrt{x^{2} - 1})$ $u = ln(x+sqrt(x^2-1))$

and:

$\int ln (x + \sqrt{x^{2} - 1}) d x = \int ln (cosh u + \sqrt{{cosh}^{2} u - 1}) \frac{d x}{d u} . d u$ $int ln(x+sqrt(x^2-1)) dx = int ln(cosh u + sqrt(cosh^2 u - 1)) dx/(du) color(white)(.) du$

$\int ln (x + \sqrt{x^{2} - 1}) d x = \int ln (cosh u + \sqrt{{sinh}^{2} u}) sinh u . d u$ $color(white)(int ln(x+sqrt(x^2-1)) dx) = int ln(cosh u + sqrt(sinh^2 u)) sinh u color(white)(.) du$

$\int ln (x + \sqrt{x^{2} - 1}) d x = \int ln (cosh u + sinh u) sinh u . d u$ $color(white)(int ln(x+sqrt(x^2-1)) dx) = int ln(cosh u + sinh u) sinh u color(white)(.) du$

$\int ln (x + \sqrt{x^{2} - 1}) d x = \int ln (e^{u}) sinh u . d u$ $color(white)(int ln(x+sqrt(x^2-1)) dx) = int ln(e^u) sinh u color(white)(.) du$

$\int ln (x + \sqrt{x^{2} - 1}) d x = \int u sinh u . d u$ $color(white)(int ln(x+sqrt(x^2-1)) dx) = int u sinh u color(white)(.) du$

$\int ln (x + \sqrt{x^{2} - 1}) d x = \int (u sinh u + cosh u) - cosh u . d u$ $color(white)(int ln(x+sqrt(x^2-1)) dx) = int (u sinh u + cosh u)-cosh u color(white)(.) du$

$\int ln (x + \sqrt{x^{2} - 1}) d x = u cosh u - sinh u + C$ $color(white)(int ln(x+sqrt(x^2-1)) dx) = u cosh u - sinh u + C$

$\int ln (x + \sqrt{x^{2} - 1}) d x = x ln (x + \sqrt{x^{2} - 1}) - \sqrt{x^{2} - 1} + C$ $color(white)(int ln(x+sqrt(x^2-1)) dx) = x ln(x+sqrt(x^2-1)) - sqrt(x^2-1) + C$

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How do you find the integral $ln (x + \sqrt{x^{2} - 1})$ $ln( x + sqrt(x^2 -1))$ ?

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Explanation:

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