What is the integral from 3 to 6 of x/(3sqrt(x^2-8)) ?

The integral from 3 to 6 of x/(3sqrt(x^2-8)) is (2sqrt(7)-1)/3

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integral from 3 to 6 of x/(3sqrt(x^2-8))

\int_{3}^{6} \frac{x}{3 x ^{2} - 8} d x

\frac{2 7 - 1}{3}

Decimal

1.43050 \dots

Solution steps

\int_{3}^{6} \frac{x}{3 x ^{2} - 8} d x

Take the constant out:

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

= \frac{1}{3} \cdot \int_{3}^{6} \frac{x}{x ^{2} - 8} d x

Apply u-substitution

: \int_{1}^{28} \frac{1}{2 u} d u

= \frac{1}{3} \cdot \int_{1}^{28} \frac{1}{2 u} d u

Take the constant out:

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

= \frac{1}{3} \cdot \frac{1}{2} \cdot \int_{1}^{28} \frac{1}{u} d u

Apply the Power Rule

: [2 u^{\frac{1}{2}}]_{1}^{28}

= \frac{1}{3} \cdot \frac{1}{2} [2 u^{\frac{1}{2}}]_{1}^{28}

Simplify

\frac{1}{3} \cdot \frac{1}{2} [2 u^{\frac{1}{2}}]_{1}^{28} : \frac{1}{6} [2 u]_{1}^{28}

= \frac{1}{6} [2 u]_{1}^{28}

Compute the boundaries:

47 - 2

= \frac{1}{6} (47 - 2)

Simplify

= \frac{2 7 - 1}{3}

\int 3 (x + 10) e^{x} d x

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

x \to 0 - lim (\frac{24}{tan ( x )})

x \to 1 + lim (x)

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

What is the integral from 3 to 6 of x/(3sqrt(x^2-8)) ?
The integral from 3 to 6 of x/(3sqrt(x^2-8)) is (2sqrt(7)-1)/3