Question

Integration of 1÷(root of x^2+a^2)= log|x+root of x^2+a^2|÷|a|

how is it possible?

Answer 1

you can see your answer

Answer 2

here is your answer-

first of all take x=atanθ

now,

$∫ \frac{1}{ \sqrt{ {x}^{2} + {a}^{2} } } dx \\ \\ x = atanθ \\ dx = {sec}^{2} θ \: dθ \\ \\ now \: put \: dx \: as \: given \: above \\ ∫ \frac{1}{ \sqrt{ ( {atanθ)}^{2} + {a}^{2} } } {sec}^{2} θ \: dθ \\ ∫ \frac{ {sec}^{2} θ dθ\: }{ \sqrt{1 + {tan}^{2} θ} } \frac{1}{a} \\ \\ \frac{1}{a} ∫ \frac{ {sec}^{2}θ }{secθ} \: dθ \\ \frac{1}{a} ∫ secθdθ \\ \\ \frac{1}{a} ln( \sec(θ) + tanθ) - - - (1)\\ put \: value \: of \: tanθ \\ \\ tanθ = \frac{x}{a} \\ secθ = \frac{ \sqrt{ {a}^{2} + {x}^{2} } }{a} (use \: formula \: {sec}^{2} x = 1 + {tan}^{2} x) \\ \\ put \: value \: of \: secθ \: and \: tanθ \: in \: eq.1 \: \\ get \: your \: answer.$