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

The integral from 0 to 5 of (3x+2)/(x^2+9) is 3/2 ln(34)-3ln(3)+2/3 arctan(5/3)

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integral from 0 to 5 of (3x+2)/(x^2+9)

\int_{0}^{5} \frac{3 x + 2}{x ^{2} + 9} d x

\frac{3}{2} ln (34) - 3 ln (3) + \frac{2}{3} arctan (\frac{5}{3})

Decimal Notation

2.68062 \dots

Solution steps

\int_{0}^{5} \frac{3 x + 2}{x ^{2} + 9} d x

Expand

\frac{3 x + 2}{x ^{2} + 9} : \frac{3 x}{x ^{2} + 9} + \frac{2}{x ^{2} + 9}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{0}^{5} \frac{3 x}{x ^{2} + 9} d x + \int_{0}^{5} \frac{2}{x ^{2} + 9} d x

\int_{0}^{5} \frac{3 x}{x ^{2} + 9} d x = \frac{3}{2} ln (34) - 3 ln (3)

\int_{0}^{5} \frac{2}{x ^{2} + 9} d x = \frac{2}{3} arctan (\frac{5}{3})

= \frac{3}{2} ln (34) - 3 ln (3) + \frac{2}{3} arctan (\frac{5}{3})

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What is the integral from 0 to 5 of (3x+2)/(x^2+9) ?
The integral from 0 to 5 of (3x+2)/(x^2+9) is 3/2 ln(34)-3ln(3)+2/3 arctan(5/3)