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

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

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

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

Solution steps

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

Use the common integral:

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

= ln x^{2} + 1 + x

Add a constant to the solution

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

y = sin (arccos (t))

\frac{d}{d x} (\frac{15}{x ^{2}})

\int_{0}^{12} \frac{1}{36 - x ^{2}} d x

\int_{0}^{\frac{π}{6}} t sin (2 t) d t

\frac{4}{s ^{2} ( s + 2 )} + \frac{3}{s + 2}

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