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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

\int \frac{3}{2 ( x - 1 )} d x = \frac{3}{2} ln ∣ x - 1 ∣

\int \frac{3 x + 5}{2 ( x ^{2} + 1 )} d x = \frac{1}{2} (\frac{3}{2} ln x^{2} + 1 + 5 arctan (x))

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

Add a constant to the solution

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

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