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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Solution steps

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

Apply Trigonometric Substitution

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

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

Apply u-substitution

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

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

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= \frac{1}{3} ln ∣ v ∣

Substitute back

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

Add a constant to the solution

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

(x^{2} + 1) \frac{d y}{d x} + 4 x (y - 1) = 0

\frac{d}{d x} (\frac{x - 8}{x + 8})

x \to 4 lim ((x - 4)^{5} - 5)

x \to 3 lim (- 1)

\frac{d}{d x} (\frac{3}{4} x^{\frac{4}{3}} - \frac{3}{8} x^{\frac{2}{3}} + 5)

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