What is the integral of 8/(xsqrt(x^2-9)) ?

The integral of 8/(xsqrt(x^2-9)) is 8/3 arcsec(1/3 x)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 8/(xsqrt(x^2-9))

\int \frac{8}{x x ^{2} - 9} d x

\frac{8}{3} arcsec (\frac{1}{3} x) + C

Solution steps

\int \frac{8}{x x ^{2} - 9} d x

Take the constant out:

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

= 8 \cdot \int \frac{1}{x x ^{2} - 9} d x

Apply Trigonometric Substitution

: \int \frac{1}{3} d u

= 8 \cdot \int \frac{1}{3} d u

Take the constant out:

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

= 8 \cdot \frac{1}{3} \cdot \int 1 d u

Integral of a constant:

\int a d x = a x

= 8 \cdot \frac{1}{3} \cdot 1 \cdot u

Substitute back

u = arcsec (\frac{1}{3} x)

= 8 \cdot \frac{1}{3} \cdot 1 \cdot arcsec (\frac{1}{3} x)

Simplify

8 \cdot \frac{1}{3} \cdot 1 \cdot arcsec (\frac{1}{3} x) : \frac{8}{3} arcsec (\frac{1}{3} x)

= \frac{8}{3} arcsec (\frac{1}{3} x)

Add a constant to the solution

= \frac{8}{3} arcsec (\frac{1}{3} x) + C

\int \frac{19 x ^{2} + 50 x - 25}{3 x ^{3} - 5 x ^{2}} d x

\int (4 \frac{sin ( 2 x )}{sin ( x )}) d x

\int 9 x^{3} + 7 3 x^{2} d x

\int \frac{x ^{4} + x ^{3} + x + 5}{x + 1} d x

\int \frac{1}{2 + sin ( θ ) + cos ( θ )} d θ

What is the integral of 8/(xsqrt(x^2-9)) ?
The integral of 8/(xsqrt(x^2-9)) is 8/3 arcsec(1/3 x)+C