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

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

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

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

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

Solution steps

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

Apply Trigonometric Substitution

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

= 2 \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

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

Apply u-substitution

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

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

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= 2 \cdot \frac{1}{2} ln ∣ v ∣

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

\int \frac{2 x + 7}{x ^{2} + 5 x - 4} d x

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\int \frac{4 x ^{2}}{1 - 16 x ^{3}} d x

\int \frac{( 2 x ^{2} )}{x ^{2} - 1} d x

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