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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from-1 to 1 of 2/(1+4t^2)

\int_{- 1}^{1} \frac{2}{1 + 4 t ^{2}} d t

2 arctan (2)

Decimal Notation

2.21429 \dots

Solution steps

\int_{- 1}^{1} \frac{2}{1 + 4 t ^{2}} d t

Take the constant out:

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

= 2 \cdot \int_{- 1}^{1} \frac{1}{1 + 4 t ^{2}} d t

Apply Integral Substitution

= 2 \cdot \int_{- 2}^{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

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

Use the common integral:

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

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

Simplify

2 \cdot \frac{1}{2} [arctan (u)]_{- 2}^{2} : [arctan (u)]_{- 2}^{2}

= [arctan (u)]_{- 2}^{2}

Compute the boundaries:

2 arctan (2)

= 2 arctan (2)

\int_{4}^{6} x (\frac{1}{x - 3}) d x

\int_{- 8}^{\infty} \frac{1}{x + 9} d x

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

\int_{0}^{- 1} x \cdot e^{x} d x

\int_{0}^{2} (x + 2)^{2} - 3 d x

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