What is the integral from 0 to 2 of (8x)/(sqrt(2x^2+1)) ?

The integral from 0 to 2 of (8x)/(sqrt(2x^2+1)) is 8

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integral from 0 to 2 of (8x)/(sqrt(2x^2+1))

\int_{0}^{2} \frac{8 x}{2 x ^{2} + 1} d x

8

Solution steps

\int_{0}^{2} \frac{8 x}{2 x ^{2} + 1} d x

Take the constant out:

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

= 8 \cdot \int_{0}^{2} \frac{x}{2 x ^{2} + 1} d x

Apply u-substitution

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

= 8 \cdot \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

= 8 \cdot \frac{1}{4} \cdot \int_{1}^{9} \frac{1}{u} d u

Apply the Power Rule

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

= 8 \cdot \frac{1}{4} [2 u^{\frac{1}{2}}]_{1}^{9}

Simplify

8 \cdot \frac{1}{4} [2 u^{\frac{1}{2}}]_{1}^{9} : 2 [2 u]_{1}^{9}

= 2 [2 u]_{1}^{9}

Compute the boundaries:

4

= 2 \cdot 4

Simplify

= 8

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What is the integral from 0 to 2 of (8x)/(sqrt(2x^2+1)) ?
The integral from 0 to 2 of (8x)/(sqrt(2x^2+1)) is 8