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

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

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integral of 3/(xsqrt(x^2+4))

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

\frac{3}{2} ln tan (\frac{1}{2} arctan (\frac{1}{2} x)) + C

Solution steps

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

Take the constant out:

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

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

Apply Trigonometric Substitution

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

= 3 \cdot \int \frac{sec ( u )}{2 tan ( u )} d u

Take the constant out:

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

= 3 \cdot \frac{1}{2} \cdot \int \frac{sec ( u )}{tan ( u )} d u

Apply u-substitution

: \int \frac{1}{v} d v

= 3 \cdot \frac{1}{2} \cdot \int \frac{1}{v} d v

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= 3 \cdot \frac{1}{2} ln ∣ v ∣

Substitute back

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

Simplify

3 \cdot \frac{1}{2} ln tan (\frac{arctan ( \frac{1}{2} x )}{2}) : \frac{3}{2} ln tan (\frac{1}{2} arctan (\frac{1}{2} x))

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

Add a constant to the solution

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

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