What is the anti-derivative of 1/ ( (xsquared) + 6 ) dx ?

Question

I

Answer 1

Find the integral of 1/(x² + 6) with respect to x.

∫{1/(x² + 6)}dx

Let
(√6)tanθ = x
(√6)sec²θ dθ= dx
6tan²θ = x²
tanθ = x/√6

∫{1/(x² + 6)}dx = ∫{1/(6tan²θ + 6)}{(√6)sec²θ dθ}
= ∫{[(√6)sec²θ]/(6sec²θ)}dθ = ∫(1/√6)dθ
= θ/√6 + C = arctan(x/√6)/√6 + C

Answer 2

42

Answer 3

∫1/√(x^2+6) dx
= ln[x+√(x^2+6)]
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Let x =√6 tan u. You can derive the result.

Answer 4

arctan(x/sqrt(6))/sqrt(6)