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

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

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

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

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

Solution steps

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

Simplify

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

\frac{\partial}{\partial x} (ln (x - 6 y))

\int 2 cos (\frac{1}{2} x) d x

\frac{\partial}{\partial y} (4 x^{3} y^{2} - 2 x y^{3})

x^{3} + 7 e^{x}

\frac{x d y}{d x} = y^{2} - 4

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