What is the integral of ((4x+3))/(4x^3+8x^2+3x) ?

The integral of ((4x+3))/(4x^3+8x^2+3x) is ln|x|-1/2 ln|2x+1|-1/2 ln|2x+3|+C

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integral of ((4x+3))/(4x^3+8x^2+3x)

\int \frac{( 4 x + 3 )}{4 x ^{3} + 8 x ^{2} + 3 x} d x

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

Solution steps

\int \frac{4 x + 3}{4 x ^{3} + 8 x ^{2} + 3 x} d x

Take the partial fraction of

\frac{4 x + 3}{4 x ^{3} + 8 x ^{2} + 3 x} : \frac{1}{x} - \frac{1}{2 x + 1} - \frac{1}{2 x + 3}

= \int \frac{1}{x} - \frac{1}{2 x + 1} - \frac{1}{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

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

\int \frac{1}{x} d x = ln ∣ x ∣

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

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

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

Add a constant to the solution

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

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What is the integral of ((4x+3))/(4x^3+8x^2+3x) ?
The integral of ((4x+3))/(4x^3+8x^2+3x) is ln|x|-1/2 ln|2x+1|-1/2 ln|2x+3|+C