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

The integral of x/((x-1)(x^2+2x+2)) is 1/5 ln|x-1|+1/10 (-ln|x^2+2x+2|+6arctan(x+1))+C

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

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

\frac{1}{5} ln ∣ x - 1 ∣ + \frac{1}{10} (- ln x^{2} + 2 x + 2 + 6 arctan (x + 1)) + C

Solution steps

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

Take the partial fraction of

\frac{x}{( x - 1 ) ( x ^{2} + 2 x + 2 )} : \frac{1}{5 ( x - 1 )} + \frac{- x + 2}{5 ( x ^{2} + 2 x + 2 )}

= \int \frac{1}{5 ( x - 1 )} + \frac{- x + 2}{5 ( x ^{2} + 2 x + 2 )} 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}{5 ( x - 1 )} d x + \int \frac{- x + 2}{5 ( x ^{2} + 2 x + 2 )} d x

\int \frac{1}{5 ( x - 1 )} d x = \frac{1}{5} ln ∣ x - 1 ∣

\int \frac{- x + 2}{5 ( x ^{2} + 2 x + 2 )} d x = \frac{1}{10} (- ln x^{2} + 2 x + 2 + 6 arctan (x + 1))

= \frac{1}{5} ln ∣ x - 1 ∣ + \frac{1}{10} (- ln x^{2} + 2 x + 2 + 6 arctan (x + 1))

Add a constant to the solution

= \frac{1}{5} ln ∣ x - 1 ∣ + \frac{1}{10} (- ln x^{2} + 2 x + 2 + 6 arctan (x + 1)) + C

\frac{d}{d x} (x \cdot e^{2 x})

n = 1 \sum \infty - 30 n

\frac{\partial}{\partial t} (cos (x t) + 3 t x^{2})

\int \frac{sin ( x )}{cos ^{2} ( x ) + cos ( x ) - 6} d x

f (x) = x^{2} - 3 x + 5, at x = 1

What is the integral of x/((x-1)(x^2+2x+2)) ?
The integral of x/((x-1)(x^2+2x+2)) is 1/5 ln|x-1|+1/10 (-ln|x^2+2x+2|+6arctan(x+1))+C