What is the integral of (2x+2)/(x^2-4x+9) ?

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

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integral of (2x+2)/(x^2-4x+9)

\int \frac{2 x + 2}{x ^{2} - 4 x + 9} d x

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

Solution steps

\int \frac{2 x + 2}{x ^{2} - 4 x + 9} d x

Apply Integration By Parts

: \frac{1}{5} arctan (\frac{1}{5} (x - 2)) (2 x + 2) - \int \frac{2}{5} arctan (\frac{1}{5} (x - 2)) d x

= \frac{1}{5} arctan (\frac{1}{5} (x - 2)) (2 x + 2) - \int \frac{2}{5} arctan (\frac{1}{5} (x - 2)) d x

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

= \frac{1}{5} arctan (\frac{1}{5} (x - 2)) (2 x + 2) - (\frac{2}{5} x arctan (\frac{1}{5} (x - 2)) - ln \frac{1}{5} (x - 2)^{2} + 1 - \frac{4}{5} arctan (\frac{1}{5} (x - 2)))

Simplify

\frac{1}{5} arctan (\frac{1}{5} (x - 2)) (2 x + 2) - (\frac{2}{5} x arctan (\frac{1}{5} (x - 2)) - ln \frac{1}{5} (x - 2)^{2} + 1 - \frac{4}{5} arctan (\frac{1}{5} (x - 2))) : \frac{6}{5} arctan (\frac{1}{5} (x - 2)) + ln \frac{1}{5} (x - 2)^{2} + 1

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

Add a constant to the solution

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

\int - t e^{- s t} d t

\int_{- 1}^{2} ∣ e^{x} - 1 ∣ d x

\frac{2}{s + 1} + \frac{32 s}{s ^{2} + 9}

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

\int y^{\frac{1}{7}} d y

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