What is the integral of 3/(x^2+3x) ?

The integral of 3/(x^2+3x) is ln|x|-ln|x+3|+C

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integral of 3/(x^2+3x)

\int \frac{3}{x ^{2} + 3 x} d x

ln ∣ x ∣ - ln ∣ x + 3 ∣ + C

Solution steps

\int \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 \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 \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 \frac{1}{3 x} d x - \int \frac{1}{3 ( x + 3 )} d x)

\int \frac{1}{3 x} d x = \frac{1}{3} ln ∣ x ∣

\int \frac{1}{3 ( x + 3 )} d x = \frac{1}{3} ln ∣ x + 3 ∣

= 3 (\frac{1}{3} ln ∣ x ∣ - \frac{1}{3} ln ∣ x + 3 ∣)

Simplify

3 (\frac{1}{3} ln ∣ x ∣ - \frac{1}{3} ln ∣ x + 3 ∣) : ln ∣ x ∣ - ln ∣ x + 3 ∣

= ln ∣ x ∣ - ln ∣ x + 3 ∣

Add a constant to the solution

= ln ∣ x ∣ - ln ∣ x + 3 ∣ + C

\int t^{2} sin^{3} (t^{3}) cos^{2} (t^{3}) d t

\int \frac{- x ^{2} - 2}{( x ^{2} + x + 1 ) ^{2}} d x

tan^{2} (x)

\int \frac{1}{( x ^{2} + 1 ) ( x ^{2} + x + 1 )} d x

What is the integral of 3/(x^2+3x) ?
The integral of 3/(x^2+3x) is ln|x|-ln|x+3|+C