What is the integral from 2 to 4 of 1/(sqrt(x^2+1)) ?

The integral from 2 to 4 of 1/(sqrt(x^2+1)) is ln(sqrt(17)+4)-ln(sqrt(5)+2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 2 to 4 of 1/(sqrt(x^2+1))

\int_{2}^{4} \frac{1}{x ^{2} + 1} d x

ln (17 + 4) - ln (5 + 2)

Solution steps

\int_{2}^{4} \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]_{2}^{4}

Compute the boundaries:

ln (17 + 4) - ln (5 + 2)

= ln (17 + 4) - ln (5 + 2)

\frac{\partial}{\partial x} (y^{7 x})

y^{^{'}} = \frac{1}{2} y

4^{c o s (3 t)}

sin (t - π)

\int \frac{3 x}{x ^{2} - x - 2} d x

What is the integral from 2 to 4 of 1/(sqrt(x^2+1)) ?
The integral from 2 to 4 of 1/(sqrt(x^2+1)) is ln(sqrt(17)+4)-ln(sqrt(5)+2)