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

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

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

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

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

Solution steps

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

Apply Trigonometric Substitution

: \int \frac{sec ( u )}{2} d u

= \int \frac{sec ( u )}{2} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int sec (u) d u

Use the common integral:

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

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

Substitute back

u = arctan (2 t)

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

Simplify

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

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

Add a constant to the solution

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

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