What is the integral from 1 to 7 of x/(x^2+9) ?

The integral from 1 to 7 of x/(x^2+9) is 1/2 ln(29/5)

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integral from 1 to 7 of x/(x^2+9)

\int_{1}^{7} \frac{x}{x ^{2} + 9} d x

\frac{1}{2} ln (\frac{29}{5})

Decimal Notation

0.87892 \dots

Solution steps

\int_{1}^{7} \frac{x}{x ^{2} + 9} d x

Apply u-substitution

: \int_{10}^{58} \frac{1}{2 u} d u

= \int_{10}^{58} \frac{1}{2 u} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int_{10}^{58} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} [ln ∣ u ∣]_{10}^{58}

Compute the boundaries:

ln (58) - ln (10)

= \frac{1}{2} (ln (58) - ln (10))

Simplify

= \frac{1}{2} ln (\frac{29}{5})

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What is the integral from 1 to 7 of x/(x^2+9) ?
The integral from 1 to 7 of x/(x^2+9) is 1/2 ln(29/5)