What is the integral of 1/(sqrt(x^2+3/4)) ?

The integral of 1/(sqrt(x^2+3/4)) is ln| 2/(sqrt(3))x+sqrt(1/3 (3+4x^2))|+C

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integral of 1/(sqrt(x^2+3/4))

\int \frac{1}{x ^{2} + \frac{3}{4}} d x

ln \frac{2}{3} x + \frac{1}{3} (3 + 4 x^{2}) + C

Solution steps

\int \frac{1}{x ^{2} + \frac{3}{4}} d x

x^{2} + \frac{3}{4} = \frac{3}{4} (\frac{x ^{2}}{\frac{3}{4}} + 1)

= \int \frac{1}{\frac{3}{4} ( \frac{x ^{2}}{\frac{3}{4}} + 1 )} d x

Simplify

\frac{1}{\frac{3}{4} ( \frac{x ^{2}}{\frac{3}{4}} + 1 )} : \frac{2}{4 x ^{2} + 3}

= \int \frac{2}{4 x ^{2} + 3} d x

Take the constant out:

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

= 2 \cdot \int \frac{1}{4 x ^{2} + 3} d x

Apply Trigonometric Substitution

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

= 2 \cdot \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

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

Use the common integral:

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

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

Substitute back

u = arctan (\frac{2}{3} x)

= 2 \cdot \frac{1}{2} ln tan (arctan (\frac{2}{3} x)) + sec (arctan (\frac{2}{3} x))

Simplify

2 \cdot \frac{1}{2} ln tan (arctan (\frac{2}{3} x)) + sec (arctan (\frac{2}{3} x)) : ln \frac{2}{3} x + \frac{1}{3} (3 + 4 x^{2})

= ln \frac{2}{3} x + \frac{1}{3} (3 + 4 x^{2})

Add a constant to the solution

= ln \frac{2}{3} x + \frac{1}{3} (3 + 4 x^{2}) + C

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What is the integral of 1/(sqrt(x^2+3/4)) ?
The integral of 1/(sqrt(x^2+3/4)) is ln| 2/(sqrt(3))x+sqrt(1/3 (3+4x^2))|+C