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How do you integrate #int dx/sqrt(x^2+2x)# using trig substitutions?

Answer 1

Isaiah Allen

The answer is #=ln(|sqrt(x^2+2x)+x+1|)+C#

Complete the square in the denominator

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

Let #x+1=secu#, #=>#, #dx=secutanudu#

Therefore, the integral is

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

#=int(secutanu du)/(sqrt(tan^2u))#

#=int(secudu)#

#=int(secu(tanu+secu)du)/(tanu+secu)#

Let #v=tanu+secu#, #=>#, #dv=(secutanu+sec^2u)du#

Therefore,

#I=int(dv)/v#

#=lnv#

#=ln(tanu+secu)#

#=ln(|sqrt((x+1)^2-1)+x+1|)+C#

#=ln(|sqrt(x^2+2x)+x+1|)+C#

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Answer 2

Ezekiel Alexander

To integrate (\int \frac{dx}{\sqrt{x^2 + 2x}}) using trigonometric substitution, first, complete the square in the denominator to make it suitable for substitution. Then, let (x + 1 = \tan(\theta)). From there, use trigonometric identities to express (\sqrt{x^2 + 2x}) in terms of (\tan(\theta)). Finally, perform the substitution and integrate with respect to (\theta). After integration, back-substitute to express the result in terms of (x).

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Answer from HIX Tutor

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