What is the integral of 1/(3xsqrt(9x^2-16)) ?

The integral of 1/(3xsqrt(9x^2-16)) is 1/12 arcsec(3/4 x)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/(3xsqrt(9x^2-16))

\int \frac{1}{3 x 9 x ^{2} - 16} d x

\frac{1}{12} arcsec (\frac{3}{4} x) + C

Solution steps

\int \frac{1}{3 x 9 x ^{2} - 16} d x

Take the constant out:

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

= \frac{1}{3} \cdot \int \frac{1}{x 9 x ^{2} - 16} d x

Apply Trigonometric Substitution

: \int \frac{1}{4} d u

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

Take the constant out:

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

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

Integral of a constant:

\int a d x = a x

= \frac{1}{3} \cdot \frac{1}{4} \cdot 1 \cdot u

Substitute back

u = arcsec (\frac{3}{4} x)

= \frac{1}{3} \cdot \frac{1}{4} \cdot 1 \cdot arcsec (\frac{3}{4} x)

Simplify

\frac{1}{3} \cdot \frac{1}{4} \cdot 1 \cdot arcsec (\frac{3}{4} x) : \frac{1}{12} arcsec (\frac{3}{4} x)

= \frac{1}{12} arcsec (\frac{3}{4} x)

Add a constant to the solution

= \frac{1}{12} arcsec (\frac{3}{4} x) + C

\int 4 x e^{3 x^{2}} d x

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

\int 1 + 5 x d x

\int e^{3 x} \cdot sin (5 x) d x

\int 2 x - cos (x) d x

What is the integral of 1/(3xsqrt(9x^2-16)) ?
The integral of 1/(3xsqrt(9x^2-16)) is 1/12 arcsec(3/4 x)+C