What is the integral of 1/(x^2-4x+9) ?

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

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integral of 1/(x^2-4x+9)

\int \frac{1}{x ^{2} - 4 x + 9} d x

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

Solution steps

\int \frac{1}{x ^{2} - 4 x + 9} d x

Complete the square

x^{2} - 4 x + 9 : (x - 2)^{2} + 5

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

Apply u-substitution

: \int \frac{1}{u ^{2} + 5} d u

= \int \frac{1}{u ^{2} + 5} d u

Apply Integral Substitution

= \int \frac{1}{5 ( v ^{2} + 1 )} d v

Take the constant out:

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

= \frac{1}{5} \cdot \int \frac{1}{v ^{2} + 1} d v

Use the common integral:

\int \frac{1}{v ^{2} + 1} d v = arctan (v)

= \frac{1}{5} arctan (v)

Substitute back

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

Add a constant to the solution

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

f (x) = 6 x - x^{2}, at x = 1

z \to 0 lim (\frac{- 1}{z})

\int \frac{x ^{3}}{x ^{4} - 7} d x

\frac{d}{d x} (⌊ \frac{3}{2} ⌋)

\frac{d}{d x} (arccos (\frac{7}{x}))

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