What is the integral from 2 to 7 of (4x)/(x^2+1) ?

The integral from 2 to 7 of (4x)/(x^2+1) is 2ln(10)

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integral from 2 to 7 of (4x)/(x^2+1)

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

2 ln (10)

Decimal

4.60517 \dots

Solution steps

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

Apply u-substitution

: \int_{5}^{50} \frac{1}{2 u} d u

= 4 \cdot \int_{5}^{50} \frac{1}{2 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}{2} \cdot \int_{5}^{50} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 4 \cdot \frac{1}{2} [ln ∣ u ∣]_{5}^{50}

Simplify

4 \cdot \frac{1}{2} [ln ∣ u ∣]_{5}^{50} : 2 [ln ∣ u ∣]_{5}^{50}

= 2 [ln ∣ u ∣]_{5}^{50}

Compute the boundaries:

ln (50) - ln (5)

= 2 (ln (50) - ln (5))

Simplify

= 2 ln (10)

\int_{y}^{x} \frac{1}{x} d x

\int_{- 2}^{4} x^{2} ∣ x - 1 ∣ d x

\int_{1}^{9} 5 u d u

\int_{- 5}^{4} \frac{1}{2} x d x

\int_{5}^{10} \frac{1}{x} d x

What is the integral from 2 to 7 of (4x)/(x^2+1) ?
The integral from 2 to 7 of (4x)/(x^2+1) is 2ln(10)