What is the integral of 1+(2x^2+x+6)/((x^2-1)(x-2)) ?

The integral of 1+(2x^2+x+6)/((x^2-1)(x-2)) is x+16/3 ln|x-2|+7/6 ln|x+1|-9/2 ln|x-1|+C

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integral of 1+(2x^2+x+6)/((x^2-1)(x-2))

\int 1 + \frac{2 x ^{2} + x + 6}{( x ^{2} - 1 ) ( x - 2 )} d x

x + \frac{16}{3} ln ∣ x - 2 ∣ + \frac{7}{6} ln ∣ x + 1 ∣ - \frac{9}{2} ln ∣ x - 1 ∣ + C

Solution steps

\int 1 + \frac{2 x ^{2} + x + 6}{( x ^{2} - 1 ) ( x - 2 )} d x

\frac{2 x ^{2} + x + 6}{( x ^{2} - 1 ) ( x - 2 )} = \frac{x ^{3} + 8}{( x ^{2} - 1 ) ( x - 2 )} - 1

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

Simplify

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

Take the partial fraction of

\frac{x ^{3} + 8}{( x - 2 ) ( x ^{2} - 1 )} : 1 + \frac{16}{3 ( x - 2 )} + \frac{7}{6 ( x + 1 )} - \frac{9}{2 ( x - 1 )}

= \int 1 + \frac{16}{3 ( x - 2 )} + \frac{7}{6 ( x + 1 )} - \frac{9}{2 ( x - 1 )} 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 1 d x + \int \frac{16}{3 ( x - 2 )} d x + \int \frac{7}{6 ( x + 1 )} d x - \int \frac{9}{2 ( x - 1 )} d x

\int 1 d x = x

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

\int \frac{7}{6 ( x + 1 )} d x = \frac{7}{6} ln ∣ x + 1 ∣

\int \frac{9}{2 ( x - 1 )} d x = \frac{9}{2} ln ∣ x - 1 ∣

= x + \frac{16}{3} ln ∣ x - 2 ∣ + \frac{7}{6} ln ∣ x + 1 ∣ - \frac{9}{2} ln ∣ x - 1 ∣

Add a constant to the solution

= x + \frac{16}{3} ln ∣ x - 2 ∣ + \frac{7}{6} ln ∣ x + 1 ∣ - \frac{9}{2} ln ∣ x - 1 ∣ + C

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x^{5} e^{2 x}

What is the integral of 1+(2x^2+x+6)/((x^2-1)(x-2)) ?
The integral of 1+(2x^2+x+6)/((x^2-1)(x-2)) is x+16/3 ln|x-2|+7/6 ln|x+1|-9/2 ln|x-1|+C