What is the integral from 4 to 5 of x/(sqrt(2x-6)) ?

The integral from 4 to 5 of x/(sqrt(2x-6)) is (2(11-5sqrt(2)))/3

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integral from 4 to 5 of x/(sqrt(2x-6))

\int_{4}^{5} \frac{x}{2 x - 6} d x

\frac{2 ( 11 - 5 2 )}{3}

Decimal

2.61928 \dots

Solution steps

\int_{4}^{5} \frac{x}{2 x - 6} d x

Apply u-substitution

: \int_{2}^{2} \frac{u ^{2} + 6}{2} d u

= \int_{2}^{2} \frac{u ^{2} + 6}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int_{2}^{2} u^{2} + 6 d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{2} (\int_{2}^{2} u^{2} d u + \int_{2}^{2} 6 d u)

\int_{2}^{2} u^{2} d u = \frac{8 - 2 2}{3}

\int_{2}^{2} 6 d u = 12 - 62

= \frac{1}{2} (\frac{8 - 2 2}{3} + 12 - 62)

Simplify

\frac{1}{2} (\frac{8 - 2 2}{3} + 12 - 62) : \frac{2 ( 11 - 5 2 )}{3}

= \frac{2 ( 11 - 5 2 )}{3}

\int_{0}^{1} 5 x^{4} + 3 x^{2} + 1 d x

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\int_{1}^{2} \frac{4 ( ln ( x ) ) ^{3}}{x} d x

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\int_{1}^{2} x (2 x^{2} - 1)^{2} d x

What is the integral from 4 to 5 of x/(sqrt(2x-6)) ?
The integral from 4 to 5 of x/(sqrt(2x-6)) is (2(11-5sqrt(2)))/3