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

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

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

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

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

Solution steps

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

Simplify

= \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

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

\frac{d}{d t} (\frac{2 t - 1}{2 t + 1})

\frac{2}{s ^{5}}

x \to 0 lim (sin (x^{12}))

\frac{d}{d x} (x arcsec (x^{3}))

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