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

The integral of 1/(xsqrt(9x^2-9)) is 1/3 arctan(sqrt(x^2-1))+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(9x^2-9))

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

\frac{1}{3} arctan (x^{2} - 1) + C

Solution steps

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

Apply Integration By Parts

: \frac{1}{3 x ^{2} - 1} - \int - \frac{2 x ^{2} - 1}{3 x ( x ^{2} - 1 ) x ^{2} - 1} d x

= \frac{1}{3 x ^{2} - 1} - \int - \frac{2 x ^{2} - 1}{3 x ( x ^{2} - 1 ) x ^{2} - 1} d x

\int - \frac{2 x ^{2} - 1}{3 x ( x ^{2} - 1 ) x ^{2} - 1} d x = - \frac{1}{3} (- \frac{1}{x ^{2} - 1} + arctan (x^{2} - 1))

= \frac{1}{3 x ^{2} - 1} - (- \frac{1}{3} (- \frac{1}{x ^{2} - 1} + arctan (x^{2} - 1)))

Simplify

\frac{1}{3 x ^{2} - 1} - (- \frac{1}{3} (- \frac{1}{x ^{2} - 1} + arctan (x^{2} - 1))) : \frac{1}{3} arctan (x^{2} - 1)

= \frac{1}{3} arctan (x^{2} - 1)

Add a constant to the solution

= \frac{1}{3} arctan (x^{2} - 1) + C

\int x^{3} 1 + 36 x^{2} d x

\int (e^{- \frac{x}{5}} + 5 x^{4}) d x

h (t) = (t + 4)^{\frac{2}{3}} (2 t^{2} - 4)^{3}

x \to 7 lim (sec (\frac{π x}{6}))

\frac{d y}{d x} = 7 x + y

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