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Given,

$f(x)$ $:=$ $\sqrt{x^2 + 1} \over x$,

with $x > 0,$

I wondered which kind of integration would be the most clever one.

Following the advice of Gilbert Strang, I would try to substitute $x = \sin u,$ but in the end, this would lead me to the integration of $\csc u,$ and that's not an easy way to do it, I guess. Another approach would be to substitute $u = \sqrt{x^2 + 1}.$ This substitution works much more straight-forward, but it includes a lot of smaller calculations that could cost me time in an exam. So, do you know another approach that works with a clever trick and goes much faster?

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Julian
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2 Answers2

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Let $x=\tan y\implies\sqrt{1+x^2}=\sec y,dx=\sec^2y\ dy$

$$\int\dfrac{\sqrt{1+x^2}}x\ dx=\int\dfrac{dy}{\cos^2y\sin y}=\int\dfrac{\sin ydy}{\cos^2y(1- \cos^2y)}$$

Put $\cos y=u$

So, we can start directly with $x^2+1=\dfrac1{u^2}$ for

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

lab bhattacharjee
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Let $x=\tan t, dx=\sec^2t dt$ to get

$\displaystyle\int\frac{\sec t}{\tan t}\sec^2 t dt =\int\frac{\sec t}{\tan t}\big(\tan^2 t+1\big)dt =\int (\sec t \tan t+\csc t) dt$

$\displaystyle=\sec t +\ln|\csc t-\cot t|+C=\sqrt{x^2+1}+\ln\left|\frac{\sqrt{x^2+1}}{x}-\frac{1}{x}\right|+C$

Alternate method:

Let $x=\cot t, dx =-\csc^2t dt$ to get $\displaystyle\int (\sec t)(-\csc^2 t)dt$.

Then use integration by parts with $u=\sec t, dv=-\csc^2 t dt$ to get

$\displaystyle (\sec t)(\cot t)-\int(\cot t)(\sec t\tan t) dt=\csc t-\int\sec t dt$

$\displaystyle=\csc t-\ln|\sec t+\tan t|+C=\sqrt{x^2+1}-\ln\left|\frac{\sqrt{x^2+1}}{x}+\frac{1}{x}\right|+C$

user84413
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