What is the integral of (x^2)/(x^2-6x+25) ?

The integral of (x^2)/(x^2-6x+25) is 1/4 (12ln| 1/16 (x-3)^2+1|-7arctan(1/4 (x-3)))+x+C

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integral of (x^2)/(x^2-6x+25)

\int \frac{x ^{2}}{x ^{2} - 6 x + 25} d x

\frac{1}{4} (12 ln \frac{1}{16} (x - 3)^{2} + 1 - 7 arctan (\frac{1}{4} (x - 3))) + x + C

Solution steps

\int \frac{x ^{2}}{x ^{2} - 6 x + 25} d x

\frac{x ^{2}}{x ^{2} - 6 x + 25} = \frac{6 x - 25}{x ^{2} - 6 x + 25} + 1

= \int \frac{6 x - 25}{x ^{2} - 6 x + 25} + 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{6 x - 25}{x ^{2} - 6 x + 25} d x + \int 1 d x

\int \frac{6 x - 25}{x ^{2} - 6 x + 25} d x = \frac{1}{4} (12 ln \frac{1}{16} (x - 3)^{2} + 1 - 7 arctan (\frac{1}{4} (x - 3)))

\int 1 d x = x

= \frac{1}{4} (12 ln \frac{1}{16} (x - 3)^{2} + 1 - 7 arctan (\frac{1}{4} (x - 3))) + x

Add a constant to the solution

= \frac{1}{4} (12 ln \frac{1}{16} (x - 3)^{2} + 1 - 7 arctan (\frac{1}{4} (x - 3))) + x + C

x \to - \infty lim (e^{4 x})

\int_{4}^{8} \frac{x}{x ^{2} + 4 x + 20} d x

\frac{\partial}{\partial x} (x^{x})

x \to \infty lim (9.35 e^{(- 0.4 x)} + 5.52)

x \to - 2 lim (\frac{x ^{2} + 2}{x + 2})

What is the integral of (x^2)/(x^2-6x+25) ?
The integral of (x^2)/(x^2-6x+25) is 1/4 (12ln| 1/16 (x-3)^2+1|-7arctan(1/4 (x-3)))+x+C