What is the integral from 3 to 5 of x/(x^2-4) ?

The integral from 3 to 5 of x/(x^2-4) is 1/2 (ln(21)-ln(5))

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integral from 3 to 5 of x/(x^2-4)

\int_{3}^{5} \frac{x}{x ^{2} - 4} d x

\frac{1}{2} (ln (21) - ln (5))

Decimal

0.71754 \dots

Solution steps

\int_{3}^{5} \frac{x}{x ^{2} - 4} d x

Apply u-substitution

: \int_{5}^{21} \frac{1}{2 u} d u

= \int_{5}^{21} \frac{1}{2 u} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int_{5}^{21} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} [ln ∣ u ∣]_{5}^{21}

Compute the boundaries:

ln (21) - ln (5)

= \frac{1}{2} (ln (21) - ln (5))

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

\int_{3}^{10} \frac{1}{x} d x

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

\int_{1}^{2} (f (t) + f^{- 1} (t)) d t

\int_{1}^{2} x^{2} - 4 x + 3 d x

What is the integral from 3 to 5 of x/(x^2-4) ?
The integral from 3 to 5 of x/(x^2-4) is 1/2 (ln(21)-ln(5))