What is the integral of 2/(xsqrt(9+x^2)) ?

The integral of 2/(xsqrt(9+x^2)) is 2/3 ln|tan(1/2 arctan(1/3 x))|+C

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integral of 2/(xsqrt(9+x^2))

\int \frac{2}{x 9 + x ^{2}} d x

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

Solution steps

\int \frac{2}{x 9 + x ^{2}} 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 9 + x ^{2}} d x

Apply Trigonometric Substitution

: \int \frac{sec ( u )}{3 tan ( u )} d u

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

Apply u-substitution

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

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

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= 2 \cdot \frac{1}{3} ln ∣ v ∣

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

\int \frac{6 x + 5}{( 3 x ^{2} + 5 x ) ^{6}} d x

f (x) = \frac{( 1 - 4 x )}{3 + x}

\frac{d}{d x} (1 0^{9 - x^{2}})

t^{2} sin (2 t)

x y^{^{'}} = y + 4 x^{2} sin (x), y (π) = 0

What is the integral of 2/(xsqrt(9+x^2)) ?
The integral of 2/(xsqrt(9+x^2)) is 2/3 ln|tan(1/2 arctan(1/3 x))|+C