What is the integral from 1 to 1.6 of (2x)/(x^2-4) ?

The integral from 1 to 1.6 of (2x)/(x^2-4) is -0.73396…

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integral from 1 to 1.6 of (2x)/(x^2-4)

\int_{1}^{1.6} \frac{2 x}{x ^{2} - 4} d x

- 0.73396 \dots

Solution steps

\int_{1}^{1.6} \frac{2 x}{x ^{2} - 4} d x

Take the constant out:

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

= 2 \cdot \int_{1}^{1.6} \frac{x}{x ^{2} - 4} d x

Apply u-substitution

: \int_{- 3}^{- 1.44} \frac{1}{2 u} d u

= 2 \cdot \int_{- 3}^{- 1.44} \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}^{- 1.44} \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}^{- 1.44}

Simplify

2 \cdot \frac{1}{2} [ln ∣ u ∣]_{- 3}^{- 1.44} : [ln ∣ u ∣]_{- 3}^{- 1.44}

= [ln ∣ u ∣]_{- 3}^{- 1.44}

Compute the boundaries:

- 0.73396 \dots

= - 0.73396 \dots

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

\int_{0}^{1} (x^{2} + y^{2}) d x

\int_{1}^{5} 3 x + 1 d x

\int_{0}^{\frac{15}{16}} 1 + 16 x d x

\int_{1}^{\infty} x^{1} d x

What is the integral from 1 to 1.6 of (2x)/(x^2-4) ?
The integral from 1 to 1.6 of (2x)/(x^2-4) is -0.73396…