What is the integral from-2 to 0 of (-2x)/(1+x^2) ?

The integral from-2 to 0 of (-2x)/(1+x^2) is ln(5)

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integral from-2 to 0 of (-2x)/(1+x^2)

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

ln (5)

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

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

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - 2 (- \int_{1}^{5} \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 (- \frac{1}{2} \cdot \int_{1}^{5} \frac{1}{u} d u)

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - 2 (- \frac{1}{2} [ln ∣ u ∣]_{1}^{5})

Simplify

- 2 (- \frac{1}{2} [ln ∣ u ∣]_{1}^{5}) : [ln ∣ u ∣]_{1}^{5}

= [ln ∣ u ∣]_{1}^{5}

Compute the boundaries:

ln (5)

= ln (5)

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\int_{5}^{\infty} x e^{- 5 x} d x

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

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

What is the integral from-2 to 0 of (-2x)/(1+x^2) ?
The integral from-2 to 0 of (-2x)/(1+x^2) is ln(5)