What is the integral from 1 to 5 of 3/(x^2+3x) ?

The integral from 1 to 5 of 3/(x^2+3x) is ln(5)-ln(2)

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integral from 1 to 5 of 3/(x^2+3x)

\int_{1}^{5} \frac{3}{x ^{2} + 3 x} d x

ln (5) - ln (2)

Decimal Notation

0.91629 \dots

Solution steps

\int_{1}^{5} \frac{3}{x ^{2} + 3 x} d x

Take the constant out:

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

= 3 \cdot \int_{1}^{5} \frac{1}{x ^{2} + 3 x} d x

Take the partial fraction of

\frac{1}{x ^{2} + 3 x} : \frac{1}{3 x} - \frac{1}{3 ( x + 3 )}

= 3 \cdot \int_{1}^{5} \frac{1}{3 x} - \frac{1}{3 ( x + 3 )} d x

Apply the Sum Rule:

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

= 3 (\int_{1}^{5} \frac{1}{3 x} d x - \int_{1}^{5} \frac{1}{3 ( x + 3 )} d x)

\int_{1}^{5} \frac{1}{3 x} d x = \frac{1}{3} ln (5)

\int_{1}^{5} \frac{1}{3 ( x + 3 )} d x = \frac{1}{3} ln (2)

= 3 (\frac{1}{3} ln (5) - \frac{1}{3} ln (2))

Simplify

3 (\frac{1}{3} ln (5) - \frac{1}{3} ln (2)) : ln (5) - ln (2)

= ln (5) - ln (2)

\frac{\partial}{\partial y} (x^{5} + x^{2} y + x + 2)

\int \frac{1}{x} - e^{- x} d x

\int \frac{1}{4 x + 6 x} d x

n = 1 \sum \infty \frac{1}{2 n - 1}

x \frac{d y}{d x} = \frac{x ^{3}}{y} - \frac{5 x ^{5}}{y} - \frac{1}{x y}

What is the integral from 1 to 5 of 3/(x^2+3x) ?
The integral from 1 to 5 of 3/(x^2+3x) is ln(5)-ln(2)