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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

\int 1 d x = x

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

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

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

Add a constant to the solution

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

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