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

The integral of 1/((x-3)(x^2+1)) is 1/10 ln|x-3|+1/10 (-1/2 ln|x^2+1|-3arctan(x))+C

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

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

\frac{1}{10} ln ∣ x - 3 ∣ + \frac{1}{10} (- \frac{1}{2} ln x^{2} + 1 - 3 arctan (x)) + C

Solution steps

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

Take the partial fraction of

\frac{1}{( x - 3 ) ( x ^{2} + 1 )} : \frac{1}{10 ( x - 3 )} + \frac{- x - 3}{10 ( x ^{2} + 1 )}

= \int \frac{1}{10 ( x - 3 )} + \frac{- x - 3}{10 ( 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}{10 ( x - 3 )} d x + \int \frac{- x - 3}{10 ( x ^{2} + 1 )} d x

\int \frac{1}{10 ( x - 3 )} d x = \frac{1}{10} ln ∣ x - 3 ∣

\int \frac{- x - 3}{10 ( x ^{2} + 1 )} d x = \frac{1}{10} (- \frac{1}{2} ln x^{2} + 1 - 3 arctan (x))

= \frac{1}{10} ln ∣ x - 3 ∣ + \frac{1}{10} (- \frac{1}{2} ln x^{2} + 1 - 3 arctan (x))

Add a constant to the solution

= \frac{1}{10} ln ∣ x - 3 ∣ + \frac{1}{10} (- \frac{1}{2} ln x^{2} + 1 - 3 arctan (x)) + C

\frac{\partial}{\partial y} (x y e^{8 y})

x \to 2 + lim (\frac{x + 7}{x ^{2} - 4})

x \to 0 lim (\frac{1 - cos ( 9 x )}{x})

f (x) e^{6} + ln (4)

y = \frac{27}{x ^{2} + 9}, at x = 2

What is the integral of 1/((x-3)(x^2+1)) ?
The integral of 1/((x-3)(x^2+1)) is 1/10 ln|x-3|+1/10 (-1/2 ln|x^2+1|-3arctan(x))+C