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

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

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

\int \frac{1}{( 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)) + C

Solution steps

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

Take the partial fraction of

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

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

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

Add a constant to the solution

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

y^{^{''}} - 5 y^{^{'}} - 6 y = 12 e^{2} t

\frac{\partial}{\partial x} (y e^{3 x})

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

\frac{\partial}{\partial x} (2)

t \cdot y^{^{'}} + 4 y = 8 t

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