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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/(sqrt(x^2+1))

integral

\frac{1}{x ^{2} + 1}

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

Solution steps

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

Use the common integral:

\int \frac{1}{x ^{2} + 1} d x = ln x^{2} + 1 + x

= ln x^{2} + 1 + x

Add a constant to the solution

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

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

x^{3} (e^{- x})^{2}

\frac{1}{x ( k + x )}

sec (x) \cdot tan (x) + 5

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

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