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

The integral from 3 to 5 of x/(x^2-3) is 1/2 ln(11/3)

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

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

\frac{1}{2} ln (\frac{11}{3})

Decimal

0.64964 \dots

Solution steps

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

Apply u-substitution

: \int_{6}^{22} \frac{1}{2 u} d u

= \int_{6}^{22} \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_{6}^{22} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} [ln ∣ u ∣]_{6}^{22}

Compute the boundaries:

ln (22) - ln (6)

= \frac{1}{2} (ln (22) - ln (6))

Simplify

= \frac{1}{2} ln (\frac{11}{3})

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\int 10 sin^{6} (x) cos^{3} (x) d x

e^{5 x}

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What is the integral from 3 to 5 of x/(x^2-3) ?
The integral from 3 to 5 of x/(x^2-3) is 1/2 ln(11/3)