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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

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

\int \frac{- x - 2}{5 ( x ^{2} + 1 )} 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))

Add a constant to the solution

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

\int (4 z + 1)^{2} d z

\frac{d}{d x} (x n x)

\frac{\partial}{\partial x} (4^{x})

x \to - 2 lim (4 - x^{2})

d er i v a t i v e \frac{sin ( x ) - 1}{sin ( x ) + 4}

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