What is the integral from-1 to 1 of 8/(x^2+4) ?

The integral from-1 to 1 of 8/(x^2+4) is 8arctan(1/2)

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 8/(x^2+4)

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

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

Decimal

3.70918 \dots

Solution steps

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

Take the constant out:

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

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

Apply Integral Substitution

= 8 \cdot \int_{- \frac{1}{2}}^{\frac{1}{2}} \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_{- \frac{1}{2}}^{\frac{1}{2}} \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)]_{- \frac{1}{2}}^{\frac{1}{2}}

Simplify

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

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

Compute the boundaries:

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

= 4 \cdot 2 arctan (\frac{1}{2})

Simplify

= 8 arctan (\frac{1}{2})

\int_{x^{2}}^{x^{7}} (2 t - 1)^{3} d t

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

\int_{- l n (6)}^{l n (6)} cosh (x) d x

\int_{- 5}^{5} 5 + 25 - x^{2} d x

\int_{0}^{7} g (x) d x

What is the integral from-1 to 1 of 8/(x^2+4) ?
The integral from-1 to 1 of 8/(x^2+4) is 8arctan(1/2)