What is the integral of x/((x^2-1)^{1/2)} ?

The integral of x/((x^2-1)^{1/2)} is (x^2)/((x^2-1)^{1/2)}-1/((x^2-1)^{1/2)}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of x/((x^2-1)^{1/2)}

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

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

Solution steps

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

Apply Integration By Parts

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

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

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

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

Add a constant to the solution

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

2 z

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

\frac{d y}{d x}, y = x^{π}

t y^{^{'}} + t^{2} + t y - y = 0

What is the integral of x/((x^2-1)^{1/2)} ?
The integral of x/((x^2-1)^{1/2)} is (x^2)/((x^2-1)^{1/2)}-1/((x^2-1)^{1/2)}+C