What is the integral from 0 to 6 of x/(sqrt(9+2x^2)) ?

The integral from 0 to 6 of x/(sqrt(9+2x^2)) is 3

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integral from 0 to 6 of x/(sqrt(9+2x^2))

\int_{0}^{6} \frac{x}{9 + 2 x ^{2}} d x

3

Solution steps

\int_{0}^{6} \frac{x}{9 + 2 x ^{2}} d x

Apply u-substitution

= \int_{9}^{81} \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_{9}^{81} \frac{1}{u} d u

Apply the Power Rule

= \frac{1}{4} [2 u^{\frac{1}{2}}]_{9}^{81}

Apply radical rule:

= \frac{1}{4} [2 u]_{9}^{81}

Compute the boundaries:

12

= \frac{1}{4} \cdot 12

Simplify

= 3

What is the integral from 0 to 6 of x/(sqrt(9+2x^2)) ?
The integral from 0 to 6 of x/(sqrt(9+2x^2)) is 3