What is the integral of-1/(2xsqrt(x^2+9)) ?

The integral of-1/(2xsqrt(x^2+9)) is -1/6 ln|tan(1/2 arctan(1/3 x))|+C

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integral of-1/(2xsqrt(x^2+9))

\int - \frac{1}{2 x x ^{2} + 9} d x

- \frac{1}{6} ln tan (\frac{1}{2} arctan (\frac{1}{3} x)) + C

Solution steps

\int - \frac{1}{2 x x ^{2} + 9} d x

Take the constant out:

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

= - \frac{1}{2} \cdot \int \frac{1}{x x ^{2} + 9} d x

Apply Trigonometric Substitution

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

= - \frac{1}{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

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

Apply u-substitution

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

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

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= - \frac{1}{2} \cdot \frac{1}{3} ln ∣ v ∣

Substitute back

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

Simplify

- \frac{1}{2} \cdot \frac{1}{3} ln tan (\frac{arctan ( \frac{1}{3} x )}{2}) : - \frac{1}{6} ln tan (\frac{1}{2} arctan (\frac{1}{3} x))

= - \frac{1}{6} ln tan (\frac{1}{2} arctan (\frac{1}{3} x))

Add a constant to the solution

= - \frac{1}{6} ln tan (\frac{1}{2} arctan (\frac{1}{3} x)) + C

e^{x^{5}} + lo g_{4} (π)

n \to \infty lim (\frac{a}{n})

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

x \to 5 - lim (\frac{∣ x - 5 ∣}{x - 5})

\frac{d}{d x} (\frac{x ^{2}}{7 x + 4})

What is the integral of-1/(2xsqrt(x^2+9)) ?
The integral of-1/(2xsqrt(x^2+9)) is -1/6 ln|tan(1/2 arctan(1/3 x))|+C