What is the integral from 0 to 2 of (6x)/((1+x^2)^2) ?

The integral from 0 to 2 of (6x)/((1+x^2)^2) is 12/5

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integral from 0 to 2 of (6x)/((1+x^2)^2)

\int_{0}^{2} \frac{6 x}{( 1 + x ^{2} ) ^{2}} d x

\frac{12}{5}

Decimal Notation

2.4

Solution steps

\int_{0}^{2} \frac{6 x}{( 1 + x ^{2} ) ^{2}} d x

Take the constant out:

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

= 6 \cdot \int_{0}^{2} \frac{x}{( 1 + x ^{2} ) ^{2}} d x

Apply u-substitution

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

= 6 \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

= 6 \cdot \frac{1}{2} \cdot \int_{1}^{5} \frac{1}{u ^{2}} d u

Apply the Power Rule

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

= 6 \cdot \frac{1}{2} [- \frac{1}{u}]_{1}^{5}

Simplify

6 \cdot \frac{1}{2} [- \frac{1}{u}]_{1}^{5} : 3 [- \frac{1}{u}]_{1}^{5}

= 3 [- \frac{1}{u}]_{1}^{5}

Compute the boundaries:

\frac{4}{5}

= 3 \cdot \frac{4}{5}

Simplify

= \frac{12}{5}

\frac{\partial}{\partial y} (y - x^{3})

\frac{d}{d x} (\frac{( 1 - x )}{x})

y^{^{''}} + 6 y^{^{'}} + 9 y = t^{- 2} e^{- 3 t}

y^{^{'}} = \frac{3}{( x + 2 ) ^{2}}

7^{x + a r c c o s (x)}

What is the integral from 0 to 2 of (6x)/((1+x^2)^2) ?
The integral from 0 to 2 of (6x)/((1+x^2)^2) is 12/5