# Using the method of trigonometric show your users that ∫ dx/√(x^2+2x) = In |x + 1 + √(x^2+2x)| + constant

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You should complete the square to denominator using the following formula such that:

`(a+b)^2 = a^2 + 2ab + b^2`

`x^2 + 2x = a^2 + 2ab => {(x^2=a^2 => a=x),(2x = 2xb => b=1):}`

Hence, you need to add 1 to complete the square and subtract 1 to preserve the equation such that:

`x^2 + 2x + 1 - 1 = (x + 1)^2 - 1`

Hence, substituting `(x + 1)^2 - 1` for `x^2 + 2x` yields:

`int 1/sqrt(x^2+2x)dx = int 1/(sqrt((x + 1)^2 - 1))dx`

You need to use the following substitution such that:

`x+1 = t => dx = dt`

Changing the variable yields:

`int 1/(sqrt(t^2 - 1))dt`

You should notice that you may write `1/(sqrt(t^2 - 1)) = (arccosh t)'` such that:

`int 1/(sqrt(t^2 - 1))dt = int (arccosh t)' = arccosh t + c`

You may write the inverse function `arccosh t` using logarithms such that:

`arccosh t = ln(t + sqrt(t^2-1))`

`int 1/(sqrt(t^2 - 1))dt = ln(t + sqrt(t^2-1))+ c `

Substituting back `x+1` for `t` yields:

`int 1/sqrt(x^2+2x) dx = ln|x+1 + sqrt((x+1)^2-1)|+ c`

**Hence, evaluating the given indefinite integral using the inverse of hyperbolic function `cosh x` yields `int 1/sqrt(x^2+2x) dx = ln|x+1 + sqrt((x+1)^2-1)|+ c.` **