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

The integral from 3 to 4 of x/(x-3) is diverges

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

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

diverges

Solution steps

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

Apply u-substitution

: \int_{0}^{1} \frac{u + 3}{u} d u

= \int_{0}^{1} \frac{u + 3}{u} d u

Expand

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

= \int_{0}^{1} 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_{0}^{1} 1 d u + \int_{0}^{1} \frac{3}{u} d u

\int_{0}^{1} 1 d u = 1

\int_{0}^{1} \frac{3}{u} d u =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

\int ln (\frac{x}{1 + x}) d x

\frac{d}{d x} (2 x^{2} cos (x) cot (x))

\frac{d y}{d x}, x y - 8 = 0

\frac{d}{d x} (3 a + b x^{3})

f (x) = 5 x

What is the integral from 3 to 4 of x/(x-3) ?
The integral from 3 to 4 of x/(x-3) is diverges