What is the integral from 0 to 1 of 1/(1+4x^2) ?

The integral from 0 to 1 of 1/(1+4x^2) is 1/2 arctan(2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to 1 of 1/(1+4x^2)

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

\frac{1}{2} arctan (2)

Decimal Notation

0.55357 \dots

Solution steps

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

Apply Integral Substitution

= \int_{0}^{2} \frac{1}{2 ( 1 + u ^{2} )} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int_{0}^{2} \frac{1}{1 + u ^{2}} d u

Use the common integral:

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

= \frac{1}{2} [arctan (u)]_{0}^{2}

Compute the boundaries:

arctan (2)

= \frac{1}{2} arctan (2)

\int_{0}^{6} 4 x - x^{2} d x

\int_{0}^{2} 8 - x^{3} d x

\int_{1}^{3} 4 x^{3} - 3 x^{2} + 1 d x

\int_{0}^{4} (x 16 - x^{2}) d x

\int_{0}^{5} (x + 1)^{2} d x

What is the integral from 0 to 1 of 1/(1+4x^2) ?
The integral from 0 to 1 of 1/(1+4x^2) is 1/2 arctan(2)