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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

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

\frac{π}{4} - arctan (x)

Solution steps

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

Use the common integral:

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

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

Compute the boundaries:

\frac{π}{4} - arctan (x)

= \frac{π}{4} - arctan (x)

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