What is the integral of 1/(x^2sqrt(9x^2-4)) ?

The integral of 1/(x^2sqrt(9x^2-4)) is (sqrt(9x^2-4))/(4x)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/(x^2sqrt(9x^2-4))

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

\frac{9 x ^{2} - 4}{4 x} + C

Solution steps

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

Apply Trigonometric Substitution

: \int \frac{3}{4 sec ( u )} d u

= \int \frac{3}{4 sec ( u )} d u

Take the constant out:

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

= \frac{3}{4} \cdot \int \frac{1}{sec ( u )} d u

Rewrite using trig identities

= \frac{3}{4} \cdot \int cos (u) d u

Use the common integral:

\int cos (u) d u = sin (u)

= \frac{3}{4} sin (u)

Substitute back

u = arcsec (\frac{3}{2} x)

= \frac{3}{4} sin (arcsec (\frac{3}{2} x))

Simplify

\frac{3}{4} sin (arcsec (\frac{3}{2} x)) : \frac{9 x ^{2} - 4}{4 x}

= \frac{9 x ^{2} - 4}{4 x}

Add a constant to the solution

= \frac{9 x ^{2} - 4}{4 x} + C

\int x^{2} e^{x^{3} + 2} d x

\frac{d}{d x} (- \frac{4}{1 - 4 x})

\frac{1}{x - 1}, x = 7

y = x^{3} - 2 x^{2} + x + 7

(arcsin (x))^{^{'}}

What is the integral of 1/(x^2sqrt(9x^2-4)) ?
The integral of 1/(x^2sqrt(9x^2-4)) is (sqrt(9x^2-4))/(4x)+C