What is the integral from 0 to 4 of ((5x)/(x^2+9)) ?

The integral from 0 to 4 of ((5x)/(x^2+9)) is 5ln(5)-5ln(3)

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integral from 0 to 4 of ((5x)/(x^2+9))

\int_{0}^{4} (\frac{5 x}{x ^{2} + 9}) d x

5 ln (5) - 5 ln (3)

Decimal Notation

2.55412 \dots

Solution steps

\int_{0}^{4} \frac{5 x}{x ^{2} + 9} d x

Take the constant out:

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

= 5 \cdot \int_{0}^{4} \frac{x}{x ^{2} + 9} d x

Apply u-substitution

: \int_{9}^{25} \frac{1}{2 u} d u

= 5 \cdot \int_{9}^{25} \frac{1}{2 u} d u

Take the constant out:

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

= 5 \cdot \frac{1}{2} \cdot \int_{9}^{25} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 5 \cdot \frac{1}{2} [ln ∣ u ∣]_{9}^{25}

Simplify

5 \cdot \frac{1}{2} [ln ∣ u ∣]_{9}^{25} : \frac{5}{2} [ln ∣ u ∣]_{9}^{25}

= \frac{5}{2} [ln ∣ u ∣]_{9}^{25}

Compute the boundaries:

2 ln (5) - 2 ln (3)

= \frac{5}{2} (2 ln (5) - 2 ln (3))

Simplify

= 5 ln (5) - 5 ln (3)

\int (x + 6) (x - 7) d x

\frac{\partial}{\partial x} (\frac{x + q}{x + r})

\frac{d}{d x} ((2 x + x) (x^{2} - x))

sec^{2} (θ)

\frac{d}{d x} (\frac{cos ( x )}{x ^{6}})

What is the integral from 0 to 4 of ((5x)/(x^2+9)) ?
The integral from 0 to 4 of ((5x)/(x^2+9)) is 5ln(5)-5ln(3)