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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

\int \frac{1}{2 x} d x = \frac{1}{2} ln ∣ x ∣

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

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

Add a constant to the solution

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

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