What is the integral from 2 to 4 of (2x)/(x^2-1) ?

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

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integral from 2 to 4 of (2x)/(x^2-1)

\int_{2}^{4} \frac{2 x}{x ^{2} - 1} d x

ln (5)

Solution steps

\int_{2}^{4} \frac{2 x}{x ^{2} - 1} d x

Take the constant out:

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

= 2 \cdot \int_{2}^{4} \frac{x}{x ^{2} - 1} d x

Apply u-substitution

: \int_{3}^{15} \frac{1}{2 u} d u

= 2 \cdot \int_{3}^{15} \frac{1}{2 u} d u

Take the constant out:

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

= 2 \cdot \frac{1}{2} \cdot \int_{3}^{15} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 2 \cdot \frac{1}{2} [ln ∣ u ∣]_{3}^{15}

Simplify

2 \cdot \frac{1}{2} [ln ∣ u ∣]_{3}^{15} : [ln ∣ u ∣]_{3}^{15}

= [ln ∣ u ∣]_{3}^{15}

Compute the boundaries:

ln (15) - ln (3)

= ln (15) - ln (3)

Simplify

= ln (5)

\int_{0}^{\infty} e

\int_{- 3}^{7} \frac{2 x + 6}{5} d x

\int_{- 5}^{5} - x^{2} + 25 d x

\int_{1}^{2} \frac{8}{x ^{3}} d x

\int_{1}^{2} \frac{x ^{4} + 3 x ^{7}}{x ^{5}} d x

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