Integrate sqrt(1 + x^2) / x: Trig Substitution

[annie122]

how do u integrate sqrt(1 + x^2) / x??
i reduced this to sec^3(u)/tan(u) but how do i go from here??

 

[polygamma]

Re: trig substitution

You can do it without trig substitution.

$$ \int \frac{\sqrt{1+x^{2}}}{x} \ dx $$

Let $u^{2} = 1+x^{2}$.

Then

$$\int \frac{\sqrt{1+x^{2}}}{x} \ dx = \int \frac{u^{2}}{x^{2}} \ du = \int \frac{u^{2}}{u^{2}-1} \ du$$

$$ = \int \Big( 1 + \frac{1}{u^{2}-1} \Big) \ du = \int \Big( 1 + \frac{1}{2(u-1)} - \frac{1}{2(u+1)} \Big) \ du$$

$$ = u + \frac{\ln(u-1)}{2} - \frac{\ln(u+1)}{2} + C$$

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

 

[HallsofIvy]

how do u integrate sqrt(1 + x^2) / x??
i reduced this to sec^3(u)/tan(u) but how do i go from here??

I would be inclined to do this like Random Variable but since you got this far, sec(u)= 1/cos(u) and tan(u)= sin(u)/cos(u) so that $$\frac{sec^3(u)}{tan(u)}= \frac{1}{cos^3(u)}\frac{cos(u)}{sin(u)}= \frac{1}{sin(u)cos^2(u)}$$which has sine to an odd power. We can multiply both numerator and denominator by sin(x) to get $$\frac{sin(x)}{sin^2(x)cos^2(x)}= \frac{sin(x)}{(1- cos^2(x))cos^2(x)}$$

Let v= cos(x) so that dv= -sin(x)dx and the integrand becomes $$\frac{-dv}{(1- v^2)v^2}$$ which can be integrated using "partial fractions".

 

[soroban]

Hello, Yuuki

I found an approach.
I'm sure someone will have a better way.

$$\int \frac{\sqrt{1 + x^2}}{x}\,dx$$

$$\text{Let }\,x \,=\,\tan\theta \quad\Rightarrow\quad dx \,=\,\sec^2\!\theta\,d\theta$$

$$\text{I reduced this to: }\:\int \frac{\sec^3\!\theta}{\tan\theta}\,d\theta$$ . Good!

$$\text{Multiply by }\frac{\tan\theta}{\tan\theta}:\;\int\frac{\sec^3 \!\theta\tan\theta}{\tan^2\!\theta}\,d\theta$$

. . $$=\;\int\frac{\sec^3\!\theta\tan \theta\,d\theta}{\sec^2\!\theta-1} \;=\;\int\frac{\sec^2\!\theta}{\sec^2\!\theta-1}( \sec\theta\tan\theta\,d\theta)$$$$\text{Let }u \,=\,\sec\theta \quad\Rightarrow\quad du \,=\,\sec\theta \tan\theta\,d\theta$$

$$\text{We have: }\:\int \frac{u^2}{u^2-1}\,du \;=\;\int\left(1 + \frac{1}{u^2-1}\right)du$$

. . . . . . $$=\;u + \frac{1}{2}\ln\left|\frac{u-1}{u+1}\right|+C$$

Now back-substitute . . .

 

[Krizalid1]

Another solution:

$$\int {\frac{{\sqrt {1 + {x^2}} }}{x}\,dx} = \int {\frac{{1 + {x^2}}}{{x\sqrt {1 + {x^2}} }}\,dx} = \int {\frac{{dx}}{{x\sqrt {1 + {x^2}} }}} + \int {\frac{x}{{\sqrt {1 + {x^2}} }}\,dx} ,$$

second integral is easy, but if we want to avoid partial fractions for the first one, put $x=\dfrac1t$ and the integral becomes
$$- \int {\frac{{dt}}{{\sqrt {1 + {t^2}} }}} = - \ln \left| {t + \sqrt {1 + {t^2}} } \right| + {k_1}.$$

Finally,

$$\int {\frac{{\sqrt {1 + {x^2}} }}{x}\,dx} = - \ln \left| {\frac{1}{x} + \sqrt {1 + \frac{1}{{{x^2}}}} } \right| + \sqrt {1 + {x^2}} + k.$$