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I need help with the proof of the table of integral that: $$\int \frac{\sqrt{a^2 + u^2}}{u} du = \sqrt{a^2 + u^2} + a \ln \left|\frac{\sqrt{a^2 + u^2} - a}{u}\right| + c$$

Must solve using trigonometric substitution.

I understand that you have to use $a^2\tan(\theta)$ as a substitution for $u$.

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  • $\begingroup$ Is this what you meant? If not, you can use MathJax to change it yourself. $\endgroup$ – Toby Mak Feb 4 '19 at 13:32
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    $\begingroup$ take the derivation of the right side and compare it with the left? $\endgroup$ – Bonbon Feb 4 '19 at 13:48
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    $\begingroup$ Why must you use trig substitution? Of course, to prove something in the integral table, you merely have to differentiate the right hand side and verify that it is the integrand. $\endgroup$ – GEdgar Feb 5 '19 at 1:30
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Let $u=a \tan x \tag{1}$ Then:- $u^2=a^2\tan^2 x \implies 2u \cdot du= 2a^2\tan x \cdot \sec^2x \cdot dx $

The integral reduces to :- $$ a\int \sec^2x \csc x \cdot dx $$ $$ = a\int (\tan^2x +1) \csc x \cdot dx = a\left(\int \sec x \tan x \cdot dx + \int \csc x \cdot dx \right)\\ = a\left(\sec x - \ln(\cot x + \csc x)\right) + C $$Note that from $(1)$ we have $\sec x = \frac{\sqrt{u^2+a^2}}{a}$ and $\csc x= \frac{\sqrt{u^2+a^2}}{u}$ and $\cot x = \frac{a}{u} $

Thus , by undoing substitution , we get $$\boxed{ I= \sqrt{a^2+u^2} + a\cdot \ln \left|\frac{\sqrt{a^2+u^2}-a}{u}\right| + C}$$

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  • $\begingroup$ Thanks for your help! $\endgroup$ – Christian Martinez Feb 5 '19 at 16:37

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