What is the integral of (3x^2-4x+5)/((x-1)(x^2+1)) ?

The integral of (3x^2-4x+5)/((x-1)(x^2+1)) is 2ln|x-1|+1/2 ln|x^2+1|-3arctan(x)+C

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integral of (3x^2-4x+5)/((x-1)(x^2+1))

\int \frac{3 x ^{2} - 4 x + 5}{( x - 1 ) ( x ^{2} + 1 )} d x

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

Solution steps

\int \frac{3 x ^{2} - 4 x + 5}{( x - 1 ) ( x ^{2} + 1 )} d x

Take the partial fraction of

\frac{3 x ^{2} - 4 x + 5}{( x - 1 ) ( x ^{2} + 1 )} : \frac{2}{x - 1} + \frac{x - 3}{x ^{2} + 1}

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

\int \frac{2}{x - 1} d x = 2 ln ∣ x - 1 ∣

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

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

Add a constant to the solution

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

\frac{d}{d x} ((sin (2 x))^{x})

\int 3 (5^{1 - 2 x}) d x

\frac{d}{d x} (40 x)

x \to - 2 lim (3 x - 1)

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

What is the integral of (3x^2-4x+5)/((x-1)(x^2+1)) ?
The integral of (3x^2-4x+5)/((x-1)(x^2+1)) is 2ln|x-1|+1/2 ln|x^2+1|-3arctan(x)+C