What is the integral of 1/(9x^2-18x+72) ?

The integral of 1/(9x^2-18x+72) is 1/(9sqrt(7))arctan((x-1)/(sqrt(7)))+C

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integral of 1/(9x^2-18x+72)

\int \frac{1}{9 x ^{2} - 18 x + 72} d x

\frac{1}{9 7} arctan (\frac{x - 1}{7}) + C

Solution steps

\int \frac{1}{9 x ^{2} - 18 x + 72} d x

Complete the square

9 x^{2} - 18 x + 72 : 9 (x - 1)^{2} + 63

= \int \frac{1}{9 ( x - 1 ) ^{2} + 63} d x

Apply u-substitution

: \int \frac{1}{9 u ^{2} + 63} d u

= \int \frac{1}{9 u ^{2} + 63} d u

Apply Integral Substitution

= \int \frac{1}{9 7 ( 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}{9 7} \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}{9 7} arctan (v)

Substitute back

= \frac{1}{9 7} arctan (\frac{3}{3 7} (x - 1))

Simplify

\frac{1}{9 7} arctan (\frac{3}{3 7} (x - 1)) : \frac{1}{9 7} arctan (\frac{x - 1}{7})

= \frac{1}{9 7} arctan (\frac{x - 1}{7})

Add a constant to the solution

= \frac{1}{9 7} arctan (\frac{x - 1}{7}) + C

\int - 4 ln (2 x + 1) d x

(y^{- 1} + y^{- 2}) (2 y^{- 3} - 6 y^{- 4})

x \to \infty lim (\frac{1}{x \cdot lo g _{e} ( x )})

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

x \to \infty lim (1 + \frac{a}{x})

What is the integral of 1/(9x^2-18x+72) ?
The integral of 1/(9x^2-18x+72) is 1/(9sqrt(7))arctan((x-1)/(sqrt(7)))+C