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

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

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

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

1

Solution steps

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

Apply u-substitution

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

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

Apply the Power Rule

: [2 u^{\frac{1}{2}}]_{1}^{9}

= \frac{1}{4} [2 u^{\frac{1}{2}}]_{1}^{9}

Apply radical rule:

= \frac{1}{4} [2 u]_{1}^{9}

Compute the boundaries:

4

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

Simplify

= 1

\frac{d}{d x} (\frac{e ^{x}}{4 x ^{2}})

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

y^{^{''}} - 8 y^{^{'}} + 9 y = 0, y (0) = 0, y^{^{'}} (0) = 9

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

x \to - 1 lim (4 x^{2} - 8 x + 3)

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