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

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

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

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

ln ∣ x - 1 ∣ + arctan (x) + C

Solution steps

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

Expand

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

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

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

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

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

Simplify

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

= ln ∣ x - 1 ∣ + arctan (x)

Add a constant to the solution

= ln ∣ x - 1 ∣ + arctan (x) + C

\int \frac{x + 5 x ^{7}}{x ^{3}} d x

\frac{ln ( x )}{x ^{2}}

x ln (x)

\int \frac{1}{81 + 64 x ^{2}} d x

\int \frac{4 x ^{2}}{4 - 9 x ^{6}} d x

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