How do you prove the integral formula $\int \frac{d x}{\sqrt{x^{2} + a^{2}}} = ln (x + \sqrt{x^{2} + a^{2}}) + C$ $intdx/(sqrt(x^2+a^2)) = ln(x+sqrt(x^2+a^2))+ C$ ?

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How do you prove the integral formula $\int \frac{d x}{\sqrt{x^{2} + a^{2}}} = ln (x + \sqrt{x^{2} + a^{2}}) + C$ $intdx/(sqrt(x^2+a^2)) = ln(x+sqrt(x^2+a^2))+ C$ ?

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Answer 1 · 2014-09-11T00:52:43

Let $x = a tan θ$ $x=atan theta$ . $\Rightarrow d x = a {sec}^{2} θ d θ$ $Rightarrow dx=asec^2theta d theta$
So, we can write
$\int \frac{d x}{\sqrt{x^{2} + a^{2}}} = \int \frac{a {sec}^{2} θ}{\sqrt{a^{2} ({tan}^{2} θ + 1)}} d θ$ $int{dx}/{sqrt{x^2+a^2}}=int{asec^2theta}/{sqrt{a^2(tan^2theta+1)}}d theta$
by ${tan}^{2} θ + 1 = {sec}^{2} θ$ $tan^2theta+1=sec^2theta$ ,
$= \int sec θ d θ = ln | sec θ + tan θ | + C$ $=intsec theta d theta=ln|sec theta+tan theta|+C$
since $tan θ = \frac{x}{a}$ $tan theta=x/a$ and $sec θ = \frac{\sqrt{x^{2} + a^{2}}}{a}$ $sec theta={sqrt{x^2+a^2}}/a$ ,
$= ln ∣ ∣ ∣ \frac{\sqrt{x^{2} + a^{2}} + x}{a} ∣ ∣ ∣ + C_{1}$ $=ln|{sqrt{x^2+a^2}+x}/a|+C_1$
by the log property $ln {\frac{A}{B}} = ln A - ln B$ $ln{A/B}=lnA-lnB$ ,
$= ln ∣ ∣ \sqrt{x^{2} + a^{2}} + x ∣ ∣ - ln | a | + C_{1}$ $=ln|sqrt{x^2+a^2}+x|-ln|a|+C_1$
by setting $C = ln | a | + C_{1}$ $C=ln|a|+C_1$ ,
$= ln ∣ ∣ x + \sqrt{x^{2} + a^{2}} ∣ ∣ + C$ $=ln|x+sqrt{x^2+a^2}|+C$

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How do you prove the integral formula $\int \frac{d x}{\sqrt{x^{2} + a^{2}}} = ln (x + \sqrt{x^{2} + a^{2}}) + C$ $intdx/(sqrt(x^2+a^2)) = ln(x+sqrt(x^2+a^2))+ C$ ?

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How do you prove the integral formula $\int \frac{d x}{\sqrt{x^{2} + a^{2}}} = ln (x + \sqrt{x^{2} + a^{2}}) + C$ $intdx/(sqrt(x^2+a^2)) = ln(x+sqrt(x^2+a^2))+ C$ ?