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

The integral of 1/((x^2+1)^{1/2)} is ln|x+sqrt(x^2+1)|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

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

ln x + x^{2} + 1 + C

Solution steps

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

Apply Integration By Parts

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

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

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

Simplify

= ln x + x^{2} + 1

Add a constant to the solution

= ln x + x^{2} + 1 + C

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