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

The integral of ((x^2+x))/((x-3)(x^2+1)) is 6/5 ln|x-3|-1/10 ln|x^2+1|+2/5 arctan(x)+C

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

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

\frac{6}{5} ln ∣ x - 3 ∣ - \frac{1}{10} ln x^{2} + 1 + \frac{2}{5} arctan (x) + C

Solution steps

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

Expand

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

\int \frac{x ^{2}}{( x - 3 ) ( x ^{2} + 1 )} d x = \frac{9}{10} ln ∣ x - 3 ∣ + \frac{1}{10} (\frac{1}{2} ln x^{2} + 1 + 3 arctan (x))

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

= \frac{9}{10} ln ∣ x - 3 ∣ + \frac{1}{10} (\frac{1}{2} ln x^{2} + 1 + 3 arctan (x)) + \frac{3}{10} ln ∣ x - 3 ∣ + \frac{1}{10} (- \frac{3}{2} ln x^{2} + 1 + arctan (x))

Simplify

\frac{9}{10} ln ∣ x - 3 ∣ + \frac{1}{10} (\frac{1}{2} ln x^{2} + 1 + 3 arctan (x)) + \frac{3}{10} ln ∣ x - 3 ∣ + \frac{1}{10} (- \frac{3}{2} ln x^{2} + 1 + arctan (x)) : \frac{6}{5} ln ∣ x - 3 ∣ - \frac{1}{10} ln x^{2} + 1 + \frac{2}{5} arctan (x)

= \frac{6}{5} ln ∣ x - 3 ∣ - \frac{1}{10} ln x^{2} + 1 + \frac{2}{5} arctan (x)

Add a constant to the solution

= \frac{6}{5} ln ∣ x - 3 ∣ - \frac{1}{10} ln x^{2} + 1 + \frac{2}{5} arctan (x) + C

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

\frac{d}{d x} (\frac{x ^{2}}{20})

\frac{d}{d x} (3 x - 4 x)

\int_{1}^{2} \frac{x}{1 + 4 x ^{2}} d x

(- \frac{3}{7}, 0), (0, 6)

What is the integral of ((x^2+x))/((x-3)(x^2+1)) ?
The integral of ((x^2+x))/((x-3)(x^2+1)) is 6/5 ln|x-3|-1/10 ln|x^2+1|+2/5 arctan(x)+C