What is the integral from 2 to 6 of x/(x^2-3) ?

The integral from 2 to 6 of x/(x^2-3) is 1/2 ln(33)

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integral from 2 to 6 of x/(x^2-3)

\int_{2}^{6} \frac{x}{x ^{2} - 3} d x

\frac{1}{2} ln (33)

Solution steps

\int_{2}^{6} \frac{x}{x ^{2} - 3} d x

Apply u-substitution

: \int_{1}^{33} \frac{1}{2 u} d u

= \int_{1}^{33} \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_{1}^{33} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} [ln ∣ u ∣]_{1}^{33}

Compute the boundaries:

ln (33)

= \frac{1}{2} ln (33)

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

\int_{e^{- 1}}^{e^{1}} ∣ ln (x) ∣ d x

\int_{1}^{3} (1 + ln (x)) x^{x} d x

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

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

What is the integral from 2 to 6 of x/(x^2-3) ?
The integral from 2 to 6 of x/(x^2-3) is 1/2 ln(33)