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

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

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

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

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

Solution steps

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

Expand

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

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

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

Add a constant to the solution

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

\frac{d}{d x} (ln (3 x + 4))

x \to 6 + lim (x 6 - x)

x \to 0 lim ((x) sin (\frac{1}{x}))

x \to - 7 lim (\frac{x ^{2} - 3}{7 - x})

8 (x - \frac{1}{x})^{3}

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