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

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

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

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

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

Solution steps

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

Apply Trigonometric Substitution

: \int sec (v) d v

= \int sec (v) d v

Use the common integral:

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

= ln ∣ tan (v) + sec (v) ∣

Substitute back

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

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

Simplify

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

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

Add a constant to the solution

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

f (x) = (x^{2} + 3)^{3} \cdot 4 x^{2}

x \to 0 lim (\frac{( ln ( x ) ) ^{2}}{x})

y^{^{'}} = (x + y)^{2}

\frac{1 + csc ( x ^{3} )}{1 - cot ( x ^{3} )}

\frac{d}{d x} (24)

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