What is the integral of 8/(3x^3+7x^2+4x) ?

The integral of 8/(3x^3+7x^2+4x) is 8(1/4 ln|x|-ln|x+1|+3/4 ln|3x+4|)+C

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integral of 8/(3x^3+7x^2+4x)

\int \frac{8}{3 x ^{3} + 7 x ^{2} + 4 x} d x

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

Solution steps

\int \frac{8}{3 x ^{3} + 7 x ^{2} + 4 x} d x

Take the constant out:

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

= 8 \cdot \int \frac{1}{3 x ^{3} + 7 x ^{2} + 4 x} d x

Take the partial fraction of

\frac{1}{3 x ^{3} + 7 x ^{2} + 4 x} : \frac{1}{4 x} - \frac{1}{x + 1} + \frac{9}{4 ( 3 x + 4 )}

= 8 \cdot \int \frac{1}{4 x} - \frac{1}{x + 1} + \frac{9}{4 ( 3 x + 4 )} 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

= 8 (\int \frac{1}{4 x} d x - \int \frac{1}{x + 1} d x + \int \frac{9}{4 ( 3 x + 4 )} d x)

\int \frac{1}{4 x} d x = \frac{1}{4} ln ∣ x ∣

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

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

= 8 (\frac{1}{4} ln ∣ x ∣ - ln ∣ x + 1 ∣ + \frac{3}{4} ln ∣ 3 x + 4 ∣)

Add a constant to the solution

= 8 (\frac{1}{4} ln ∣ x ∣ - ln ∣ x + 1 ∣ + \frac{3}{4} ln ∣ 3 x + 4 ∣) + C

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What is the integral of 8/(3x^3+7x^2+4x) ?
The integral of 8/(3x^3+7x^2+4x) is 8(1/4 ln|x|-ln|x+1|+3/4 ln|3x+4|)+C