Using the method of trigonometric show your users that ∫ dx/√(x^2+2x) = In |x + 1 + √(x^2+2x)| + constant
- print Print
- list Cite
Expert Answers
calendarEducator since 2011
write5,349 answers
starTop subjects are Math, Science, and Business
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.`
Related Questions
- Find the indefinite integral `int x sqrt((2x-1)) dx` using integration by substitution `int x...
- 1 Educator Answer
- Show that `tan^2 x = (1 - cos(2x))/(1 + cos(2x))`
- 1 Educator Answer
- Using the method of partial fractions show your users that ∫ dx/x^2(x-2) = -1/4In(x) + 1/2x +...
- 1 Educator Answer
- Using the method of integration for parts show that ∫xe^(x/2) dx = 2xe^(x/2) - 4e^(x/2) + constant
- 1 Educator Answer
- (e^2x+1)/e^x dxintegrate by substitution method
- 1 Educator Answer
Unlock This Answer Now
Start your 48-hour free trial to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.