What is the integral from 2 to 3 of 3/(1+x^2) ?

The integral from 2 to 3 of 3/(1+x^2) is 3(arctan(3)-arctan(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 3 of 3/(1+x^2)

\int_{2}^{3} \frac{3}{1 + x ^{2}} d x

3 (arctan (3) - arctan (2))

Decimal Notation

0.42569 \dots

Solution steps

\int_{2}^{3} \frac{3}{1 + x ^{2}} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 3 \cdot \int_{2}^{3} \frac{1}{1 + x ^{2}} d x

Use the common integral:

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

= 3 [arctan (x)]_{2}^{3}

Compute the boundaries:

arctan (3) - arctan (2)

= 3 (arctan (3) - arctan (2))

\int_{0}^{1} e^{x} e^{x} + 4 d x

\int_{0}^{2 π} 2 - 2 cos (x) d x

\int_{0}^{\frac{1}{2}} 4 e^{2 x} d x

\int_{- 1}^{0} x (x^{2} - 1)^{10} d x

\int_{0}^{e} \frac{ln ( x ^{3} )}{x} d x

What is the integral from 2 to 3 of 3/(1+x^2) ?
The integral from 2 to 3 of 3/(1+x^2) is 3(arctan(3)-arctan(2))