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

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Answer 1 · 2018-03-29T13:02:31

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,

$I=int(dv)/v$

$=lnv$

$=ln(tanu+secu)$

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

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

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