What is the integral of 1/(sqrt((4+x^2))) ?

The integral of 1/(sqrt((4+x^2))) is ln(1/2 |x+sqrt(4+x^2)|)+C

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integral of 1/(sqrt((4+x^2)))

\int \frac{1}{( 4 + x ^{2} )} d x

ln (\frac{1}{2} x + 4 + x^{2}) + C

Solution steps

\int \frac{1}{4 + x ^{2}} d x

Apply Trigonometric Substitution

: \int sec (u) d u

= \int sec (u) d u

Use the common integral:

\int sec (u) d u = ln ∣ tan (u) + sec (u) ∣

= ln ∣ tan (u) + sec (u) ∣

Substitute back

u = arctan (\frac{1}{2} x)

= ln tan (arctan (\frac{1}{2} x)) + sec (arctan (\frac{1}{2} x))

Simplify

ln tan (arctan (\frac{1}{2} x)) + sec (arctan (\frac{1}{2} x)) : ln (\frac{1}{2} x + 4 + x^{2})

= ln (\frac{1}{2} x + 4 + x^{2})

Add a constant to the solution

= ln (\frac{1}{2} x + 4 + x^{2}) + C

y = \frac{2}{3} (x^{2} + 1)^{\frac{3}{2}}

y = 5 x tan (x), at x = π

f (x) = 3^{x^{5}}

x \to 0 + lim (x^{(\frac{1}{x})})

x \to 1 - lim (x + 3)

What is the integral of 1/(sqrt((4+x^2))) ?
The integral of 1/(sqrt((4+x^2))) is ln(1/2 |x+sqrt(4+x^2)|)+C