What is the integral of x/(9x^2-6x+10) ?

The integral of x/(9x^2-6x+10) is 1/54 (3ln|9x^2-6x+10|+2arctan(x-1/3))+C

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integral of x/(9x^2-6x+10)

\int \frac{x}{9 x ^{2} - 6 x + 10} d x

\frac{1}{54} (3 ln 9 x^{2} - 6 x + 10 + 2 arctan (x - \frac{1}{3})) + C

Solution steps

\int \frac{x}{9 x ^{2} - 6 x + 10} d x

Complete the square

9 x^{2} - 6 x + 10 : 9 (x - \frac{1}{3})^{2} + 9

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

Apply u-substitution

: \int \frac{3 u + 1}{3 ( 9 u ^{2} + 9 )} d u

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

Take the constant out:

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

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

Expand

\frac{3 u + 1}{9 u ^{2} + 9} : \frac{3 u}{9 u ^{2} + 9} + \frac{1}{9 u ^{2} + 9}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int \frac{3 u}{9 u ^{2} + 9} d u = \frac{1}{6} ln 9 u^{2} + 9

\int \frac{1}{9 u ^{2} + 9} d u = \frac{1}{9} arctan (u)

= \frac{1}{3} (\frac{1}{6} ln 9 u^{2} + 9 + \frac{1}{9} arctan (u))

Substitute back

u = x - \frac{1}{3}

= \frac{1}{3} (\frac{1}{6} ln 9 (x - \frac{1}{3})^{2} + 9 + \frac{1}{9} arctan (x - \frac{1}{3}))

Simplify

\frac{1}{3} (\frac{1}{6} ln 9 (x - \frac{1}{3})^{2} + 9 + \frac{1}{9} arctan (x - \frac{1}{3})) : \frac{1}{54} (3 ln 9 x^{2} - 6 x + 10 + 2 arctan (x - \frac{1}{3}))

= \frac{1}{54} (3 ln 9 x^{2} - 6 x + 10 + 2 arctan (x - \frac{1}{3}))

Add a constant to the solution

= \frac{1}{54} (3 ln 9 x^{2} - 6 x + 10 + 2 arctan (x - \frac{1}{3})) + C

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What is the integral of x/(9x^2-6x+10) ?
The integral of x/(9x^2-6x+10) is 1/54 (3ln|9x^2-6x+10|+2arctan(x-1/3))+C