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

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

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

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

\frac{2}{9}

Decimal Notation

0.22222 \dots

Solution steps

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

Apply u-substitution

: \int_{1}^{9} \frac{1}{4 u ^{2}} d u

= \int_{1}^{9} \frac{1}{4 u ^{2}} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int_{1}^{9} \frac{1}{u ^{2}} d u

Apply the Power Rule

: [- \frac{1}{u}]_{1}^{9}

= \frac{1}{4} [- \frac{1}{u}]_{1}^{9}

Compute the boundaries:

\frac{8}{9}

= \frac{1}{4} \cdot \frac{8}{9}

Simplify

= \frac{2}{9}

\int_{- 18}^{32} \frac{1}{2 x} d x

\frac{d ^{2}}{d x ^{2}} (x^{2} + 11)

\int_{0}^{1} x^{2} e^{x^{3} + 0} d x

y^{^{''}} + y = x

f (x) = 3 x^{2} + 3 x, at x = - 2

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