What is the integral from 6 to 8 of (2x)/(x-5) ?

The integral from 6 to 8 of (2x)/(x-5) is 2(2+5ln(3))

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integral from 6 to 8 of (2x)/(x-5)

\int_{6}^{8} \frac{2 x}{x - 5} d x

2 (2 + 5 ln (3))

Decimal Notation

14.98612 \dots

Solution steps

\int_{6}^{8} \frac{2 x}{x - 5} d x

Take the constant out:

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

= 2 \cdot \int_{6}^{8} \frac{x}{x - 5} d x

Apply u-substitution

: \int_{1}^{3} \frac{u + 5}{u} d u

= 2 \cdot \int_{1}^{3} \frac{u + 5}{u} d u

Expand

\frac{u + 5}{u} : 1 + \frac{5}{u}

= 2 \cdot \int_{1}^{3} 1 + \frac{5}{u} 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

= 2 (\int_{1}^{3} 1 d u + \int_{1}^{3} \frac{5}{u} d u)

\int_{1}^{3} 1 d u = 2

\int_{1}^{3} \frac{5}{u} d u = 5 ln (3)

= 2 (2 + 5 ln (3))

\frac{d}{d x} (\frac{4}{x - 7})

f (x) = (4 x - 2)^{\frac{1}{5}}, at x = 0

\int 2 x^{3} + x d x

f (x) = 2 x^{2} + 5 x - 3, at x = 1

\int \frac{sin ( 2 x )}{41 + cos ^{2} ( x )} d x

What is the integral from 6 to 8 of (2x)/(x-5) ?
The integral from 6 to 8 of (2x)/(x-5) is 2(2+5ln(3))