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

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

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

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

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

Solution steps

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

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} t)

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

Simplify

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

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

Add a constant to the solution

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

\frac{d y}{d x} = y (y + 1) x

x \to 3 + lim (\frac{x - 3}{x - 3})

\frac{d}{d x} (ln (12 - 9 x^{2} + 2 x^{4}))

30 e^{- x^{\frac{2}{3}}}

y = arcsin (x)

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