What is the integral of 1/(xsqrt(289x^2-289)) ?

The integral of 1/(xsqrt(289x^2-289)) is 1/17 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(289x^2-289))

\int \frac{1}{x 289 x ^{2} - 289} d x

\frac{1}{17} arctan (x^{2} - 1) + C

Solution steps

\int \frac{1}{x 289 x ^{2} - 289} d x

Apply Integration By Parts

: \frac{1}{17 x ^{2} - 1} - \int - \frac{2 x ^{2} - 1}{17 x ( x ^{2} - 1 ) x ^{2} - 1} d x

= \frac{1}{17 x ^{2} - 1} - \int - \frac{2 x ^{2} - 1}{17 x ( x ^{2} - 1 ) x ^{2} - 1} d x

\int - \frac{2 x ^{2} - 1}{17 x ( x ^{2} - 1 ) x ^{2} - 1} d x = - \frac{1}{17} (- \frac{1}{x ^{2} - 1} + arctan (x^{2} - 1))

= \frac{1}{17 x ^{2} - 1} - (- \frac{1}{17} (- \frac{1}{x ^{2} - 1} + arctan (x^{2} - 1)))

Simplify

\frac{1}{17 x ^{2} - 1} - (- \frac{1}{17} (- \frac{1}{x ^{2} - 1} + arctan (x^{2} - 1))) : \frac{1}{17} arctan (x^{2} - 1)

= \frac{1}{17} arctan (x^{2} - 1)

Add a constant to the solution

= \frac{1}{17} arctan (x^{2} - 1) + C

\frac{\partial}{\partial x} (- x y - x^{4} - y^{4})

\frac{\partial}{\partial x} ((e^{x^{2}} + e^{y^{2}})^{2})

\int_{0}^{π} sin^{2} (4 x) d x

\frac{d}{d x} (k y (x))

\int_{0}^{1} 24 + 976 e^{- 0.1 x} d x

What is the integral of 1/(xsqrt(289x^2-289)) ?
The integral of 1/(xsqrt(289x^2-289)) is 1/17 arctan(sqrt(x^2-1))+C