What is the integral of 4/(x^2(2x+3)) ?

The integral of 4/(x^2(2x+3)) is 4(-2/9 ln|x|-1/(3x)+2/9 ln|2x+3|)+C

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integral of 4/(x^2(2x+3))

\int \frac{4}{x ^{2} ( 2 x + 3 )} d x

4 (- \frac{2}{9} ln ∣ x ∣ - \frac{1}{3 x} + \frac{2}{9} ln ∣ 2 x + 3 ∣) + C

Solution steps

\int \frac{4}{x ^{2} ( 2 x + 3 )} d x

Take the constant out:

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

= 4 \cdot \int \frac{1}{x ^{2} ( 2 x + 3 )} d x

Take the partial fraction of

\frac{1}{x ^{2} ( 2 x + 3 )} : - \frac{2}{9 x} + \frac{1}{3 x ^{2}} + \frac{4}{9 ( 2 x + 3 )}

= 4 \cdot \int - \frac{2}{9 x} + \frac{1}{3 x ^{2}} + \frac{4}{9 ( 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

= 4 (- \int \frac{2}{9 x} d x + \int \frac{1}{3 x ^{2}} d x + \int \frac{4}{9 ( 2 x + 3 )} d x)

\int \frac{2}{9 x} d x = \frac{2}{9} ln ∣ x ∣

\int \frac{1}{3 x ^{2}} d x = - \frac{1}{3 x}

\int \frac{4}{9 ( 2 x + 3 )} d x = \frac{2}{9} ln ∣ 2 x + 3 ∣

= 4 (- \frac{2}{9} ln ∣ x ∣ - \frac{1}{3 x} + \frac{2}{9} ln ∣ 2 x + 3 ∣)

Add a constant to the solution

= 4 (- \frac{2}{9} ln ∣ x ∣ - \frac{1}{3 x} + \frac{2}{9} ln ∣ 2 x + 3 ∣) + C

What is the integral of 4/(x^2(2x+3)) ?
The integral of 4/(x^2(2x+3)) is 4(-2/9 ln|x|-1/(3x)+2/9 ln|2x+3|)+C