What is the integral of 1/(sqrt(4x^2+5)) ?

The integral of 1/(sqrt(4x^2+5)) is 1/2 ln| 2/(sqrt(5))x+sqrt(1/5 (5+4x^2))|+C

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integral of 1/(sqrt(4x^2+5))

\int \frac{1}{4 x ^{2} + 5} d x

\frac{1}{2} ln \frac{2}{5} x + \frac{1}{5} (5 + 4 x^{2}) + C

Solution steps

\int \frac{1}{4 x ^{2} + 5} 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 (\frac{2}{5} x)

= \frac{1}{2} ln tan (arctan (\frac{2}{5} x)) + sec (arctan (\frac{2}{5} x))

Simplify

\frac{1}{2} ln tan (arctan (\frac{2}{5} x)) + sec (arctan (\frac{2}{5} x)) : \frac{1}{2} ln \frac{2}{5} x + \frac{1}{5} (5 + 4 x^{2})

= \frac{1}{2} ln \frac{2}{5} x + \frac{1}{5} (5 + 4 x^{2})

Add a constant to the solution

= \frac{1}{2} ln \frac{2}{5} x + \frac{1}{5} (5 + 4 x^{2}) + C

x \to 0 lim (cos (π) x)

\int 3 x^{- \frac{2}{3}} d x

\int e^{2 x} (sin (3 x)) d x

\int 4 x^{2} + 7 x - 4 d x

f (x) = e^{x} - x^{8}

What is the integral of 1/(sqrt(4x^2+5)) ?
The integral of 1/(sqrt(4x^2+5)) is 1/2 ln| 2/(sqrt(5))x+sqrt(1/5 (5+4x^2))|+C