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

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

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

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

- \frac{1}{4} ln ∣ x ∣ + \frac{1}{16} ln ∣ x + 2 ∣ + \frac{3}{16} ln ∣ x - 2 ∣ + \frac{1}{4 x} + C

Solution steps

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

Expand

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

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

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

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

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

Simplify

- \frac{1}{4} ln ∣ x ∣ + \frac{1}{8} ln ∣ x + 2 ∣ + \frac{1}{8} ln ∣ x - 2 ∣ + \frac{1}{4 x} - \frac{1}{16} ln ∣ x + 2 ∣ + \frac{1}{16} ln ∣ x - 2 ∣ : - \frac{1}{4} ln ∣ x ∣ + \frac{1}{16} ln ∣ x + 2 ∣ + \frac{3}{16} ln ∣ x - 2 ∣ + \frac{1}{4 x}

= - \frac{1}{4} ln ∣ x ∣ + \frac{1}{16} ln ∣ x + 2 ∣ + \frac{3}{16} ln ∣ x - 2 ∣ + \frac{1}{4 x}

Add a constant to the solution

= - \frac{1}{4} ln ∣ x ∣ + \frac{1}{16} ln ∣ x + 2 ∣ + \frac{3}{16} ln ∣ x - 2 ∣ + \frac{1}{4 x} + C

\int (\frac{y ^{2}}{2} - y tan (x)) d x

\frac{9}{s ( s ^{2} + 9 s + 9 )}

y = (1 + 2 x)^{2}, at x = 1

\int \frac{1}{- 3 x ^{2} + 3 x + 7} d x

x \to 2 lim (\frac{1}{2 ( x - 2 )})

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