What is the integral from 0 to 1 of x/(x^2-3x) ?

The integral from 0 to 1 of x/(x^2-3x) is ln(2)-ln(3)

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integral from 0 to 1 of x/(x^2-3x)

\int_{0}^{1} \frac{x}{x ^{2} - 3 x} d x

ln (2) - ln (3)

Decimal

- 0.40546 \dots

Solution steps

\int_{0}^{1} \frac{x}{x ^{2} - 3 x} d x

Factor

\frac{x}{x ^{2} - 3 x}

= \int_{0}^{1} \frac{x}{x ( x - 3 )} d x

Cancel the common factor:

x

= \int_{0}^{1} \frac{1}{x - 3} d x

Apply u-substitution

: \int_{- 3}^{- 2} \frac{1}{u} d u

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= [ln ∣ u ∣]_{- 3}^{- 2}

Compute the boundaries:

ln (2) - ln (3)

= ln (2) - ln (3)

\int_{0}^{t} e^{- u} (t - u)^{2} d u

\int_{0}^{\frac{π}{4}} (7 sec^{2} (x)) d x

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

\int_{0}^{5} 1 + t^{2} d t

\int_{0}^{1} (\frac{1}{x ^{2} + 1}) d x

What is the integral from 0 to 1 of x/(x^2-3x) ?
The integral from 0 to 1 of x/(x^2-3x) is ln(2)-ln(3)