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

The integral from 0 to 2 of (2x)/((x^2+2)^3) is 1/9

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+2)^3)

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

\frac{1}{9}

Decimal

0.11111 \dots

Solution steps

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

Apply u-substitution

: \int_{2}^{6} \frac{1}{2 u ^{3}} d u

= 2 \cdot \int_{2}^{6} \frac{1}{2 u ^{3}} 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}^{6} \frac{1}{u ^{3}} d u

Apply the Power Rule

: [- \frac{1}{2 u ^{2}}]_{2}^{6}

= 2 \cdot \frac{1}{2} [- \frac{1}{2 u ^{2}}]_{2}^{6}

Simplify

2 \cdot \frac{1}{2} [- \frac{1}{2 u ^{2}}]_{2}^{6} : [- \frac{1}{2 u ^{2}}]_{2}^{6}

= [- \frac{1}{2 u ^{2}}]_{2}^{6}

Compute the boundaries:

\frac{1}{9}

= \frac{1}{9}

\int e^{3 x} - 18 e^{- 4 x} d x

\frac{d}{d x} (\frac{5 sin ( x )}{1 + cos ( x )})

\int_{0}^{3} (∣ 2 x - 3 ∣) d x

x \to + 4 lim (6)

∣ x - 15 ∣, \frac{x}{2}

What is the integral from 0 to 2 of (2x)/((x^2+2)^3) ?
The integral from 0 to 2 of (2x)/((x^2+2)^3) is 1/9