What is the integral of (x^2+1)/(x^3+4x^2+3x) ?

The integral of (x^2+1)/(x^3+4x^2+3x) is 1/2 ln|x^2+4x+3|+1/3 ln|x|-3/2 ln|x+1|+7/6 ln|x+3|+C

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integral of (x^2+1)/(x^3+4x^2+3x)

\int \frac{x ^{2} + 1}{x ^{3} + 4 x ^{2} + 3 x} d x

\frac{1}{2} ln x^{2} + 4 x + 3 + \frac{1}{3} ln ∣ x ∣ - \frac{3}{2} ln ∣ x + 1 ∣ + \frac{7}{6} ln ∣ x + 3 ∣ + C

Solution steps

\int \frac{x ^{2} + 1}{x ^{3} + 4 x ^{2} + 3 x} d x

Expand

\frac{x ^{2} + 1}{x ^{3} + 4 x ^{2} + 3 x} : \frac{x ^{2}}{x ^{3} + 4 x ^{2} + 3 x} + \frac{1}{x ^{3} + 4 x ^{2} + 3 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{x ^{2}}{x ^{3} + 4 x ^{2} + 3 x} d x + \int \frac{1}{x ^{3} + 4 x ^{2} + 3 x} d x

\int \frac{x ^{2}}{x ^{3} + 4 x ^{2} + 3 x} d x = \frac{1}{2} ln x^{2} + 4 x + 3 + ln ∣ x + 3 ∣ - ln ∣ x + 1 ∣

\int \frac{1}{x ^{3} + 4 x ^{2} + 3 x} d x = \frac{1}{3} ln ∣ x ∣ - \frac{1}{2} ln ∣ x + 1 ∣ + \frac{1}{6} ln ∣ x + 3 ∣

= \frac{1}{2} ln x^{2} + 4 x + 3 + ln ∣ x + 3 ∣ - ln ∣ x + 1 ∣ + \frac{1}{3} ln ∣ x ∣ - \frac{1}{2} ln ∣ x + 1 ∣ + \frac{1}{6} ln ∣ x + 3 ∣

Simplify

\frac{1}{2} ln x^{2} + 4 x + 3 + ln ∣ x + 3 ∣ - ln ∣ x + 1 ∣ + \frac{1}{3} ln ∣ x ∣ - \frac{1}{2} ln ∣ x + 1 ∣ + \frac{1}{6} ln ∣ x + 3 ∣ : \frac{1}{2} ln x^{2} + 4 x + 3 + \frac{1}{3} ln ∣ x ∣ - \frac{3}{2} ln ∣ x + 1 ∣ + \frac{7}{6} ln ∣ x + 3 ∣

= \frac{1}{2} ln x^{2} + 4 x + 3 + \frac{1}{3} ln ∣ x ∣ - \frac{3}{2} ln ∣ x + 1 ∣ + \frac{7}{6} ln ∣ x + 3 ∣

Add a constant to the solution

= \frac{1}{2} ln x^{2} + 4 x + 3 + \frac{1}{3} ln ∣ x ∣ - \frac{3}{2} ln ∣ x + 1 ∣ + \frac{7}{6} ln ∣ x + 3 ∣ + C

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What is the integral of (x^2+1)/(x^3+4x^2+3x) ?
The integral of (x^2+1)/(x^3+4x^2+3x) is 1/2 ln|x^2+4x+3|+1/3 ln|x|-3/2 ln|x+1|+7/6 ln|x+3|+C