What is the integral of (7x-3)/(7x^2-6x+3) ?

The integral of (7x-3)/(7x^2-6x+3) is 1/2 ln|49x^2-42x+21|+C

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integral of (7x-3)/(7x^2-6x+3)

\int \frac{7 x - 3}{7 x ^{2} - 6 x + 3} d x

\frac{1}{2} ln 49 x^{2} - 42 x + 21 + C

Solution steps

\int \frac{7 x - 3}{7 x ^{2} - 6 x + 3} d x

Expand

\frac{7 x - 3}{7 x ^{2} - 6 x + 3} : \frac{7 x}{7 x ^{2} - 6 x + 3} - \frac{3}{7 x ^{2} - 6 x + 3}

Apply the Sum Rule:

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

= \int \frac{7 x}{7 x ^{2} - 6 x + 3} d x - \int \frac{3}{7 x ^{2} - 6 x + 3} d x

\int \frac{7 x}{7 x ^{2} - 6 x + 3} d x = 7 (\frac{1}{14} ln 49 x^{2} - 42 x + 21 + \frac{3}{14} arctan (\frac{7}{2 3} x - \frac{3}{2}))

\int \frac{3}{7 x ^{2} - 6 x + 3} d x = \frac{3}{2} arctan (\frac{7}{2 3} x - \frac{3}{2})

= 7 (\frac{1}{14} ln 49 x^{2} - 42 x + 21 + \frac{3}{14} arctan (\frac{7}{2 3} x - \frac{3}{2})) - \frac{3}{2} arctan (\frac{7}{2 3} x - \frac{3}{2})

Simplify

7 (\frac{1}{14} ln 49 x^{2} - 42 x + 21 + \frac{3}{14} arctan (\frac{7}{2 3} x - \frac{3}{2})) - \frac{3}{2} arctan (\frac{7}{2 3} x - \frac{3}{2}) : \frac{1}{2} ln 49 x^{2} - 42 x + 21

= \frac{1}{2} ln 49 x^{2} - 42 x + 21

Add a constant to the solution

= \frac{1}{2} ln 49 x^{2} - 42 x + 21 + C

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What is the integral of (7x-3)/(7x^2-6x+3) ?
The integral of (7x-3)/(7x^2-6x+3) is 1/2 ln|49x^2-42x+21|+C