What is the integral of (x^2+2x+1)/(x^3-16x) ?

The integral of (x^2+2x+1)/(x^3-16x) is 1/2 ln(|x+4||x-4|)-7/32 ln|x+4|+9/32 ln|x-4|-1/16 ln|x|+C

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integral of (x^2+2x+1)/(x^3-16x)

\int \frac{x ^{2} + 2 x + 1}{x ^{3} - 16 x} d x

\frac{1}{2} ln (∣ x + 4 ∣ ∣ x - 4 ∣) - \frac{7}{32} ln ∣ x + 4 ∣ + \frac{9}{32} ln ∣ x - 4 ∣ - \frac{1}{16} ln ∣ x ∣ + C

Solution steps

\int \frac{x ^{2} + 2 x + 1}{x ^{3} - 16 x} d x

Expand

\frac{x ^{2} + 2 x + 1}{x ^{3} - 16 x} : \frac{x ^{2}}{x ^{3} - 16 x} + \frac{2 x}{x ^{3} - 16 x} + \frac{1}{x ^{3} - 16 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} - 16 x} d x + \int \frac{2 x}{x ^{3} - 16 x} d x + \int \frac{1}{x ^{3} - 16 x} d x

\int \frac{x ^{2}}{x ^{3} - 16 x} d x = \frac{1}{2} ln (∣ x + 4 ∣ ∣ x - 4 ∣)

\int \frac{2 x}{x ^{3} - 16 x} d x = 2 (- \frac{1}{8} ln ∣ x + 4 ∣ + \frac{1}{8} ln ∣ x - 4 ∣)

\int \frac{1}{x ^{3} - 16 x} d x = - \frac{1}{16} ln ∣ x ∣ + \frac{1}{32} ln ∣ x + 4 ∣ + \frac{1}{32} ln ∣ x - 4 ∣

= \frac{1}{2} ln (∣ x + 4 ∣ ∣ x - 4 ∣) + 2 (- \frac{1}{8} ln ∣ x + 4 ∣ + \frac{1}{8} ln ∣ x - 4 ∣) - \frac{1}{16} ln ∣ x ∣ + \frac{1}{32} ln ∣ x + 4 ∣ + \frac{1}{32} ln ∣ x - 4 ∣

Simplify

\frac{1}{2} ln (∣ x + 4 ∣ ∣ x - 4 ∣) + 2 (- \frac{1}{8} ln ∣ x + 4 ∣ + \frac{1}{8} ln ∣ x - 4 ∣) - \frac{1}{16} ln ∣ x ∣ + \frac{1}{32} ln ∣ x + 4 ∣ + \frac{1}{32} ln ∣ x - 4 ∣ : \frac{1}{2} ln (∣ x + 4 ∣ ∣ x - 4 ∣) - \frac{7}{32} ln ∣ x + 4 ∣ + \frac{9}{32} ln ∣ x - 4 ∣ - \frac{1}{16} ln ∣ x ∣

= \frac{1}{2} ln (∣ x + 4 ∣ ∣ x - 4 ∣) - \frac{7}{32} ln ∣ x + 4 ∣ + \frac{9}{32} ln ∣ x - 4 ∣ - \frac{1}{16} ln ∣ x ∣

Add a constant to the solution

= \frac{1}{2} ln (∣ x + 4 ∣ ∣ x - 4 ∣) - \frac{7}{32} ln ∣ x + 4 ∣ + \frac{9}{32} ln ∣ x - 4 ∣ - \frac{1}{16} ln ∣ x ∣ + C

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What is the integral of (x^2+2x+1)/(x^3-16x) ?
The integral of (x^2+2x+1)/(x^3-16x) is 1/2 ln(|x+4||x-4|)-7/32 ln|x+4|+9/32 ln|x-4|-1/16 ln|x|+C