How do you integrate $\int \frac{d x}{\sqrt{x^{2} + 2 x}}$ $int dx/sqrt(x^2+2x)$ using trig substitutions?

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Answer 1 · 2018-03-29T13:02:31

The answer is $= ln (∣ ∣ \sqrt{x^{2} + 2 x} + x + 1 ∣ ∣) + C$ $=ln(|sqrt(x^2+2x)+x+1|)+C$

Complete the square in the denominator

$x^{2} + 2 x = x^{2} + 2 x + 1 - 1 = {(x + 1)}^{2} - 1$ $x^2+2x=x^2+2x+1-1=(x+1)^2-1$

Let $x + 1 = sec u$ $x+1=secu$ , $\Rightarrow$ $=>$ , $d x = sec u tan u d u$ $dx=secutanudu$

Therefore, the integral is

$I = \int \frac{d x}{\sqrt{x^{2} + 2 x}} = \int \frac{d x}{\sqrt{{(x + 1)}^{2} - 1}}$ $I=int(dx)/(sqrt(x^2+2x))=int(dx)/sqrt((x+1)^2-1)$

$= \int \frac{sec u tan u d u}{\sqrt{{tan}^{2} u}}$ $=int(secutanu du)/(sqrt(tan^2u))$

$= \int (sec u d u)$ $=int(secudu)$

$= \int \frac{sec u (tan u + sec u) d u}{tan u + sec u}$ $=int(secu(tanu+secu)du)/(tanu+secu)$

Let $v = tan u + sec u$ $v=tanu+secu$ , $\Rightarrow$ $=>$ , $d v = (sec u tan u + {sec}^{2} u) d u$ $dv=(secutanu+sec^2u)du$

Therefore,

$I = \int \frac{d v}{v}$ $I=int(dv)/v$

$= ln v$ $=lnv$

$= ln (tan u + sec u)$ $=ln(tanu+secu)$

$= ln (∣ ∣ ∣ \sqrt{{(x + 1)}^{2} - 1} + x + 1 ∣ ∣ ∣) + C$ $=ln(|sqrt((x+1)^2-1)+x+1|)+C$

$= ln (∣ ∣ \sqrt{x^{2} + 2 x} + x + 1 ∣ ∣) + C$ $=ln(|sqrt(x^2+2x)+x+1|)+C$

How do you integrate ∫dx√x2+2xint dx/sqrt(x^2+2x) using trig substitutions?