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

The integral from 1 to sqrt(x of) 1/(1+t^2) is arctan(sqrt(x))-pi/4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

arctan (x) - \frac{π}{4}

Solution steps

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

Use the common integral:

\int \frac{1}{1 + t ^{2}} d t = arctan (t)

= [arctan (t)]_{1}^{x}

Compute the boundaries:

arctan (x) - \frac{π}{4}

= arctan (x) - \frac{π}{4}

e^{x} (3 x^{2} - 5 x)

\frac{d y}{d x} = \frac{( - y ^{2} )}{( 1 + y ^{2} )} \cdot (e^{- x})

\frac{d}{d x} (- 2 sin (x) - x cos (x))

\int \frac{e ^{3 x}}{x} d x

f (x) = \frac{x}{x + 1}

What is the integral from 1 to sqrt(x of) 1/(1+t^2) ?
The integral from 1 to sqrt(x of) 1/(1+t^2) is arctan(sqrt(x))-pi/4