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

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

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

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

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

Solution steps

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

Expand

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

\int \frac{x ^{2}}{( x ^{2} + 1 ) ( x - 2 )} d x = \frac{1}{5} (\frac{1}{2} ln x^{2} + 1 + 2 arctan (x)) + \frac{4}{5} ln ∣ x - 2 ∣

\int \frac{1}{( x ^{2} + 1 ) ( x - 2 )} d x = \frac{1}{5} (- \frac{1}{2} ln x^{2} + 1 - 2 arctan (x)) + \frac{1}{5} ln ∣ x - 2 ∣

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

Simplify

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

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

Add a constant to the solution

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

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(1 + e^{x}) \frac{d y}{d x} + e^{x} y = 0

\int \frac{3 x}{x + 2} d x

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