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

The integral from 0 to 2 of (2x)/((x^2+1)^2) is 4/5

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+1)^2)

\int_{0}^{2} \frac{2 x}{( x ^{2} + 1 ) ^{2}} d x

\frac{4}{5}

Decimal Notation

0.8

Solution steps

\int_{0}^{2} \frac{2 x}{( x ^{2} + 1 ) ^{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} + 1 ) ^{2}} d x

Apply u-substitution

: \int_{1}^{5} \frac{1}{2 u ^{2}} d u

= 2 \cdot \int_{1}^{5} \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_{1}^{5} \frac{1}{u ^{2}} d u

Apply the Power Rule

: [- \frac{1}{u}]_{1}^{5}

= 2 \cdot \frac{1}{2} [- \frac{1}{u}]_{1}^{5}

Simplify

2 \cdot \frac{1}{2} [- \frac{1}{u}]_{1}^{5} : [- \frac{1}{u}]_{1}^{5}

= [- \frac{1}{u}]_{1}^{5}

Compute the boundaries:

\frac{4}{5}

= \frac{4}{5}

\frac{\partial}{\partial x} (4 e^{5 x})

4 x + 6

\int x^{4} - x d x

- y^{^{''}} + y^{^{'}} + y = 0

\frac{d}{d x} (\frac{x ^{2} - 3 x + 2}{2 x ^{2} + 5 x - 1})

What is the integral from 0 to 2 of (2x)/((x^2+1)^2) ?
The integral from 0 to 2 of (2x)/((x^2+1)^2) is 4/5