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

The integral of (x^2)/((x-1)(x^2+4x+5)) is 1/10 ln|x-1|+1/20 (9ln|x^2+4x+5|-26arctan(x+2))+C

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

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

\frac{1}{10} ln ∣ x - 1 ∣ + \frac{1}{20} (9 ln x^{2} + 4 x + 5 - 26 arctan (x + 2)) + C

Solution steps

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

Take the partial fraction of

\frac{x ^{2}}{( x - 1 ) ( x ^{2} + 4 x + 5 )} : \frac{1}{10 ( x - 1 )} + \frac{9 x + 5}{10 ( x ^{2} + 4 x + 5 )}

= \int \frac{1}{10 ( x - 1 )} + \frac{9 x + 5}{10 ( x ^{2} + 4 x + 5 )} 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{1}{10 ( x - 1 )} d x + \int \frac{9 x + 5}{10 ( x ^{2} + 4 x + 5 )} d x

\int \frac{1}{10 ( x - 1 )} d x = \frac{1}{10} ln ∣ x - 1 ∣

\int \frac{9 x + 5}{10 ( x ^{2} + 4 x + 5 )} d x = \frac{1}{20} (9 ln x^{2} + 4 x + 5 - 26 arctan (x + 2))

= \frac{1}{10} ln ∣ x - 1 ∣ + \frac{1}{20} (9 ln x^{2} + 4 x + 5 - 26 arctan (x + 2))

Add a constant to the solution

= \frac{1}{10} ln ∣ x - 1 ∣ + \frac{1}{20} (9 ln x^{2} + 4 x + 5 - 26 arctan (x + 2)) + C

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What is the integral of (x^2)/((x-1)(x^2+4x+5)) ?
The integral of (x^2)/((x-1)(x^2+4x+5)) is 1/10 ln|x-1|+1/20 (9ln|x^2+4x+5|-26arctan(x+2))+C