What is the integral from 50 to 100 of x/(23+9x^2) ?

The integral from 50 to 100 of x/(23+9x^2) is 1/18 (ln(90023)-ln(22523))

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integral from 50 to 100 of x/(23+9x^2)

\int_{50}^{100} \frac{x}{23 + 9 x ^{2}} d x

\frac{1}{18} (ln (90023) - ln (22523))

Solution steps

\int_{50}^{100} \frac{x}{23 + 9 x ^{2}} d x

Apply u-substitution

: \int_{22523}^{90023} \frac{1}{18 u} d u

= \int_{22523}^{90023} \frac{1}{18 u} d u

Take the constant out:

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

= \frac{1}{18} \cdot \int_{22523}^{90023} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{18} [ln ∣ u ∣]_{22523}^{90023}

Compute the boundaries:

ln (90023) - ln (22523)

= \frac{1}{18} (ln (90023) - ln (22523))

\int_{- 4}^{7} ∣ x ∣ d x

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

\int_{1}^{16} \frac{3}{x} d x

\int_{- 2}^{1} x arctan^{2} (x) d x

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

What is the integral from 50 to 100 of x/(23+9x^2) ?
The integral from 50 to 100 of x/(23+9x^2) is 1/18 (ln(90023)-ln(22523))