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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

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

Add a constant to the solution

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

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