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

The integral from 0 to 2 of (2x)/((x^2+1)^3) is 12/25

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

\int_{0}^{2} \frac{2 x}{( x ^{2} + 1 ) ^{3}} d x

\frac{12}{25}

Solution steps

\int_{0}^{2} \frac{2 x}{( x ^{2} + 1 ) ^{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} + 1 ) ^{3}} d x

Apply u-substitution

: \int_{1}^{5} \frac{1}{2 u ^{3}} d u

= 2 \cdot \int_{1}^{5} \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_{1}^{5} \frac{1}{u ^{3}} d u

Apply the Power Rule

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

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

Simplify

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

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

Compute the boundaries:

\frac{12}{25}

= \frac{12}{25}

\frac{d}{d x} (3 x^{2} - 6 x + 1)

\frac{1}{( s - 2 ) ^{2} - 4}

y^{^{''''}} - 2 y^{^{''}} + y = 0

\frac{d}{d x} (3 cos^{2} (x) - 6 sin (x))

\frac{\partial}{\partial x} (arcsin (3 x))

What is the integral from 0 to 2 of (2x)/((x^2+1)^3) ?
The integral from 0 to 2 of (2x)/((x^2+1)^3) is 12/25