What is the integral from-2 to 2 of 8/(4+x^2) ?

The integral from-2 to 2 of 8/(4+x^2) is 2pi

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from-2 to 2 of 8/(4+x^2)

\int_{- 2}^{2} \frac{8}{4 + x ^{2}} d x

2 π

Decimal Notation

6.28318 \dots

Solution steps

\int_{- 2}^{2} \frac{8}{4 + x ^{2}} d x

Take the constant out:

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

= 8 \cdot \int_{- 2}^{2} \frac{1}{4 + x ^{2}} d x

Apply Integral Substitution

= 8 \cdot \int_{- 1}^{1} \frac{1}{2 ( u ^{2} + 1 )} d u

Take the constant out:

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

= 8 \cdot \frac{1}{2} \cdot \int_{- 1}^{1} \frac{1}{u ^{2} + 1} d u

Use the common integral:

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

= 8 \cdot \frac{1}{2} [arctan (u)]_{- 1}^{1}

Simplify

8 \cdot \frac{1}{2} [arctan (u)]_{- 1}^{1} : 4 [arctan (u)]_{- 1}^{1}

= 4 [arctan (u)]_{- 1}^{1}

Compute the boundaries:

\frac{π}{2}

= 4 \cdot \frac{π}{2}

Simplify

= 2 π

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

\int_{- 1}^{0} - x cos (nπ x) d x

\int_{2}^{6} 2 x^{2} + 4 d x

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

\int_{0}^{\frac{π}{4}} sin (θ) d θ

What is the integral from-2 to 2 of 8/(4+x^2) ?
The integral from-2 to 2 of 8/(4+x^2) is 2pi