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

The integral of (3x-4)/((x-2)(x^2+1)) is 2/5 ln|x-2|+1/5 (-ln|x^2+1|+11arctan(x))+C

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

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

\frac{2}{5} ln ∣ x - 2 ∣ + \frac{1}{5} (- ln x^{2} + 1 + 11 arctan (x)) + C

Solution steps

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

Take the partial fraction of

\frac{3 x - 4}{( x - 2 ) ( x ^{2} + 1 )} : \frac{2}{5 ( x - 2 )} + \frac{- 2 x + 11}{5 ( x ^{2} + 1 )}

= \int \frac{2}{5 ( x - 2 )} + \frac{- 2 x + 11}{5 ( 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}{5 ( x - 2 )} d x + \int \frac{- 2 x + 11}{5 ( x ^{2} + 1 )} d x

\int \frac{2}{5 ( x - 2 )} d x = \frac{2}{5} ln ∣ x - 2 ∣

\int \frac{- 2 x + 11}{5 ( x ^{2} + 1 )} d x = \frac{1}{5} (- ln x^{2} + 1 + 11 arctan (x))

= \frac{2}{5} ln ∣ x - 2 ∣ + \frac{1}{5} (- ln x^{2} + 1 + 11 arctan (x))

Add a constant to the solution

= \frac{2}{5} ln ∣ x - 2 ∣ + \frac{1}{5} (- ln x^{2} + 1 + 11 arctan (x)) + C

\int \frac{x ^{2} - 16}{x - 4} d x

\int \frac{1}{5 - y} d y

\int \frac{6 x ^{3} + 4 x ^{2} - 46 x - 30}{x ^{2} - 9} d x

\int_{0}^{\infty} e^{- 2 x + 1} d x

\frac{d v}{d t} + \frac{5.4}{59} \cdot v = - 9.81

What is the integral of (3x-4)/((x-2)(x^2+1)) ?
The integral of (3x-4)/((x-2)(x^2+1)) is 2/5 ln|x-2|+1/5 (-ln|x^2+1|+11arctan(x))+C