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

The integral from-1 to 2 of (4x)/(2x^2+1) is ln(3)

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

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

ln (3)

Decimal

1.09861 \dots

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

: \int_{3}^{9} \frac{1}{4 u} d u

= 4 \cdot \int_{3}^{9} \frac{1}{4 u} d u

Take the constant out:

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

= 4 \cdot \frac{1}{4} \cdot \int_{3}^{9} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 4 \cdot \frac{1}{4} [ln ∣ u ∣]_{3}^{9}

Simplify

4 \cdot \frac{1}{4} [ln ∣ u ∣]_{3}^{9} : [ln ∣ u ∣]_{3}^{9}

= [ln ∣ u ∣]_{3}^{9}

Compute the boundaries:

ln (3)

= ln (3)

\int_{0}^{3} (x + 1) d x

\int_{0}^{\infty} e^{t} d t

\int_{0}^{1} t^{3} (2 + t^{4})^{3} d t

\int_{\infty}^{1} \frac{1}{x + 3} d x

\int_{- 1}^{1} ∣ 2 - 4 x ∣ d x

What is the integral from-1 to 2 of (4x)/(2x^2+1) ?
The integral from-1 to 2 of (4x)/(2x^2+1) is ln(3)