What is the integral of 1/(xsqrt(9x^2+4)) ?

The integral of 1/(xsqrt(9x^2+4)) is 1/2 ln|tan((arctan(3/2 x))/2)|+C

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integral of 1/(xsqrt(9x^2+4))

\int \frac{1}{x 9 x ^{2} + 4} d x

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

Solution steps

\int \frac{1}{x 9 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{3}{2} x )}{2})

Add a constant to the solution

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

(- 6 x - ln (1 + 6 x) \cdot (1 + 6 x))^{^{'}}

n = 1 \sum \infty \frac{1}{n ^{3} + n ^{2}}

\frac{x ^{2} - 5 x + 3}{x ^{2} + 7}

f (x) = - 5 x^{2} + 2 x

f (x) = 5 x^{2} + 3

What is the integral of 1/(xsqrt(9x^2+4)) ?
The integral of 1/(xsqrt(9x^2+4)) is 1/2 ln|tan((arctan(3/2 x))/2)|+C