What is the integral from 4 to 6 of x(1/(x-3)) ?

The integral from 4 to 6 of x(1/(x-3)) is 2+3ln(3)

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integral from 4 to 6 of x(1/(x-3))

\int_{4}^{6} x (\frac{1}{x - 3}) d x

2 + 3 ln (3)

Solution steps

\int_{4}^{6} x \frac{1}{x - 3} d x

Simplify

x \frac{1}{x - 3} : \frac{x}{x - 3}

= \int_{4}^{6} \frac{x}{x - 3} d x

Apply u-substitution

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

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

Expand

\frac{u + 3}{u} : 1 + \frac{3}{u}

= \int_{1}^{3} 1 + \frac{3}{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

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

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

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

= 2 + 3 ln (3)

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\int_{5}^{10} 1

What is the integral from 4 to 6 of x(1/(x-3)) ?
The integral from 4 to 6 of x(1/(x-3)) is 2+3ln(3)