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

The integral of (3x^2)/((x-3)(x^2+1)) is 3(9/10 ln|x-3|+1/10 (1/2 ln|x^2+1|+3arctan(x)))+C

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

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

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

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Take the partial fraction of

\frac{x ^{2}}{( x - 3 ) ( x ^{2} + 1 )} : \frac{9}{10 ( x - 3 )} + \frac{x + 3}{10 ( x ^{2} + 1 )}

= 3 \cdot \int \frac{9}{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

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

\int \frac{9}{10 ( x - 3 )} d x = \frac{9}{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))

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

Add a constant to the solution

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

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What is the integral of (3x^2)/((x-3)(x^2+1)) ?
The integral of (3x^2)/((x-3)(x^2+1)) is 3(9/10 ln|x-3|+1/10 (1/2 ln|x^2+1|+3arctan(x)))+C