What is the integral of 9/(sqrt(9x^2-12x+13)) ?

The integral of 9/(sqrt(9x^2-12x+13)) is 3ln|x+sqrt((x-2/3)^2+1)-2/3 |+C

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integral of 9/(sqrt(9x^2-12x+13))

\int \frac{9}{9 x ^{2} - 12 x + 13} d x

3 ln x + (x - \frac{2}{3})^{2} + 1 - \frac{2}{3} + C

Solution steps

\int \frac{9}{9 x ^{2} - 12 x + 13} d x

Take the constant out:

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

= 9 \cdot \int \frac{1}{9 x ^{2} - 12 x + 13} d x

Complete the square

9 x^{2} - 12 x + 13 : 9 (x - \frac{2}{3})^{2} + 9

= 9 \cdot \int \frac{1}{9 ( x - \frac{2}{3} ) ^{2} + 9} d x

Apply u-substitution

: \int \frac{1}{3 u ^{2} + 1} d u

= 9 \cdot \int \frac{1}{3 u ^{2} + 1} d u

Take the constant out:

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

= 9 \cdot \frac{1}{3} \cdot \int \frac{1}{u ^{2} + 1} d u

Use the common integral:

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

= 9 \cdot \frac{1}{3} ln u^{2} + 1 + u

Substitute back

u = x - \frac{2}{3}

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

Simplify

9 \cdot \frac{1}{3} ln (x - \frac{2}{3})^{2} + 1 + x - \frac{2}{3} : 3 ln x + (x - \frac{2}{3})^{2} + 1 - \frac{2}{3}

= 3 ln x + (x - \frac{2}{3})^{2} + 1 - \frac{2}{3}

Add a constant to the solution

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

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What is the integral of 9/(sqrt(9x^2-12x+13)) ?
The integral of 9/(sqrt(9x^2-12x+13)) is 3ln|x+sqrt((x-2/3)^2+1)-2/3 |+C