What is the integral from 0 to 1 of x/(1-x) ?

The integral from 0 to 1 of x/(1-x) is diverges

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integral from 0 to 1 of x/(1-x)

\int_{0}^{1} \frac{x}{1 - x} d x

diverges

Solution steps

\int_{0}^{1} \frac{x}{1 - x} d x

Apply u-substitution

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

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

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

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

Take the constant out:

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

= - (- \int_{0}^{1} \frac{- u + 1}{u} d u)

Expand

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

= - (- \int_{0}^{1} - 1 + \frac{1}{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{1}{u} d u))

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

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

diverges

If part of the expression diverges, then entire expression diverges

= diverges

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

\int_{2}^{3} \frac{2}{x ^{3}} d x

\int_{0}^{x} \frac{1}{1 - t} d t

\int_{0}^{4} (4 x - 3 x) d x

\int_{2}^{4} \frac{2 x}{x ^{2} - 1} d x

What is the integral from 0 to 1 of x/(1-x) ?
The integral from 0 to 1 of x/(1-x) is diverges