What is the integral from 0 to 2 of (2x)/((x^2+3)^2) ?

The integral from 0 to 2 of (2x)/((x^2+3)^2) is 4/21

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to 2 of (2x)/((x^2+3)^2)

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

\frac{4}{21}

Solution steps

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

Take the constant out:

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

= 2 \cdot \int_{0}^{2} \frac{x}{( x ^{2} + 3 ) ^{2}} d x

Apply u-substitution

: \int_{3}^{7} \frac{1}{2 u ^{2}} d u

= 2 \cdot \int_{3}^{7} \frac{1}{2 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_{3}^{7} \frac{1}{u ^{2}} d u

Apply the Power Rule

: [- \frac{1}{u}]_{3}^{7}

= 2 \cdot \frac{1}{2} [- \frac{1}{u}]_{3}^{7}

Simplify

2 \cdot \frac{1}{2} [- \frac{1}{u}]_{3}^{7} : [- \frac{1}{u}]_{3}^{7}

= [- \frac{1}{u}]_{3}^{7}

Compute the boundaries:

\frac{4}{21}

= \frac{4}{21}

(14, 448), (1, y)

\int 4

x \to 0 + lim (\frac{1}{x} - \frac{1}{∣ x ∣})

\frac{d}{d x} ((1 + e^{x})^{4})

\frac{7 x}{x + 1}

What is the integral from 0 to 2 of (2x)/((x^2+3)^2) ?
The integral from 0 to 2 of (2x)/((x^2+3)^2) is 4/21