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

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

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

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

Take the partial fraction of

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

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

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

Add a constant to the solution

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

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

x \to 6 lim (\frac{x ^{2}}{x - 6})

\frac{d y}{d x} + y^{5} x = - 6 y

y^{^{''}} + 4 y^{^{'}} + 29 y = 0

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

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