What is the integral of (x^2)/(x^2-4x+13) ?

The integral of (x^2)/(x^2-4x+13) is 1/3 (6ln| 1/9 (x-2)^2+1|-5arctan(1/3 (x-2)))+x+C

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integral of (x^2)/(x^2-4x+13)

\int \frac{x ^{2}}{x ^{2} - 4 x + 13} d x

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

Solution steps

\int \frac{x ^{2}}{x ^{2} - 4 x + 13} d x

\frac{x ^{2}}{x ^{2} - 4 x + 13} = \frac{4 x - 13}{x ^{2} - 4 x + 13} + 1

= \int \frac{4 x - 13}{x ^{2} - 4 x + 13} + 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{4 x - 13}{x ^{2} - 4 x + 13} d x + \int 1 d x

\int \frac{4 x - 13}{x ^{2} - 4 x + 13} d x = \frac{1}{3} (6 ln \frac{1}{9} (x - 2)^{2} + 1 - 5 arctan (\frac{1}{3} (x - 2)))

\int 1 d x = x

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

Add a constant to the solution

= \frac{1}{3} (6 ln \frac{1}{9} (x - 2)^{2} + 1 - 5 arctan (\frac{1}{3} (x - 2))) + x + C

\int 1 + cos^{2} (t) d t

\int (x^{2} - 4 x + 6) d x

\int \frac{1}{I n x ^{2} ( I n x ) ^{2} - 1} d x

\int \frac{15}{x x ^{2} - 1} d x

What is the integral of (x^2)/(x^2-4x+13) ?
The integral of (x^2)/(x^2-4x+13) is 1/3 (6ln| 1/9 (x-2)^2+1|-5arctan(1/3 (x-2)))+x+C