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

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

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

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

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

Solution steps

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

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 x)

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

Simplify

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

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

Add a constant to the solution

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

x \to 1 lim (3 f (x) - 2 h (x))

y = 3^{c o s (2 x)} + e^{3}

\int 7 sin (\frac{θ}{3}) d θ

\frac{\partial}{\partial y} (tan (x^{2} + y))

arcsin (x^{5})

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