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

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

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

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

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

Solution steps

\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

(x + 8 y) y^{^{'}} = 2 x - y

\frac{d}{d x} (x e^{x y} cos (2 x))

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

\int_{0}^{8} y + 1 d y

x \to 0 + lim (tan (\frac{4}{x}))

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