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

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

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

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

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

Solution steps

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

Cancel the common factor:

x

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

Apply Trigonometric Substitution

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

= \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

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

Apply u-substitution

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

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

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= \frac{1}{2} ln ∣ v ∣

Substitute back

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

Add a constant to the solution

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

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