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

The integral of (x^3+4)/((x^2-1)(x^2+3x+2)) is -3/4 ln|x+1|+3/(2(x+1))+5/12 ln|x-1|+4/3 ln|x+2|+C

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

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

- \frac{3}{4} ln ∣ x + 1 ∣ + \frac{3}{2 ( x + 1 )} + \frac{5}{12} ln ∣ x - 1 ∣ + \frac{4}{3} ln ∣ x + 2 ∣ + C

Solution steps

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

Take the partial fraction of

\frac{x ^{3} + 4}{( x ^{2} - 1 ) ( x ^{2} + 3 x + 2 )} : - \frac{3}{4 ( x + 1 )} - \frac{3}{2 ( x + 1 ) ^{2}} + \frac{5}{12 ( x - 1 )} + \frac{4}{3 ( x + 2 )}

= \int - \frac{3}{4 ( x + 1 )} - \frac{3}{2 ( x + 1 ) ^{2}} + \frac{5}{12 ( x - 1 )} + \frac{4}{3 ( x + 2 )} 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{3}{4 ( x + 1 )} d x - \int \frac{3}{2 ( x + 1 ) ^{2}} d x + \int \frac{5}{12 ( x - 1 )} d x + \int \frac{4}{3 ( x + 2 )} d x

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

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

\int \frac{5}{12 ( x - 1 )} d x = \frac{5}{12} ln ∣ x - 1 ∣

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

= - \frac{3}{4} ln ∣ x + 1 ∣ - (- \frac{3}{2 ( x + 1 )}) + \frac{5}{12} ln ∣ x - 1 ∣ + \frac{4}{3} ln ∣ x + 2 ∣

Simplify

= - \frac{3}{4} ln ∣ x + 1 ∣ + \frac{3}{2 ( x + 1 )} + \frac{5}{12} ln ∣ x - 1 ∣ + \frac{4}{3} ln ∣ x + 2 ∣

Add a constant to the solution

= - \frac{3}{4} ln ∣ x + 1 ∣ + \frac{3}{2 ( x + 1 )} + \frac{5}{12} ln ∣ x - 1 ∣ + \frac{4}{3} ln ∣ x + 2 ∣ + C

\int \frac{x + e ^{x}}{x e ^{x}} d x

\int \frac{3 x}{( x - 2 ) ^{2}} d x

\int \frac{4 x + 5}{x ^{2} + 4 x + 20} d x

\int \frac{1}{1 - csc ( x )} d x

x \cdot e^{2 x}

What is the integral of (x^3+4)/((x^2-1)(x^2+3x+2)) ?
The integral of (x^3+4)/((x^2-1)(x^2+3x+2)) is -3/4 ln|x+1|+3/(2(x+1))+5/12 ln|x-1|+4/3 ln|x+2|+C