What is the integral of 6/(2x^2+3x) ?

The integral of 6/(2x^2+3x) is 6(1/3 ln|x|-1/3 ln|2x+3|)+C

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integral of 6/(2x^2+3x)

\int \frac{6}{2 x ^{2} + 3 x} d x

6 (\frac{1}{3} ln ∣ x ∣ - \frac{1}{3} ln ∣ 2 x + 3 ∣) + C

Solution steps

\int \frac{6}{2 x ^{2} + 3 x} d x

Take the constant out:

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

= 6 \cdot \int \frac{1}{2 x ^{2} + 3 x} d x

Take the partial fraction of

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

= 6 \cdot \int \frac{1}{3 x} - \frac{2}{3 ( 2 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

= 6 (\int \frac{1}{3 x} d x - \int \frac{2}{3 ( 2 x + 3 )} d x)

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

\int \frac{2}{3 ( 2 x + 3 )} d x = \frac{1}{3} ln ∣ 2 x + 3 ∣

= 6 (\frac{1}{3} ln ∣ x ∣ - \frac{1}{3} ln ∣ 2 x + 3 ∣)

Add a constant to the solution

= 6 (\frac{1}{3} ln ∣ x ∣ - \frac{1}{3} ln ∣ 2 x + 3 ∣) + C

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What is the integral of 6/(2x^2+3x) ?
The integral of 6/(2x^2+3x) is 6(1/3 ln|x|-1/3 ln|2x+3|)+C