What is the integral of (x-1)/(sqrt(3x^2-2x+5)) ?

The integral of (x-1)/(sqrt(3x^2-2x+5)) is

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integral of (x-1)/(sqrt(3x^2-2x+5))

\int \frac{x - 1}{3 x ^{2} - 2 x + 5} d x

\frac{1}{3 3} (9 x^{2} - 6 x + 15 + ln \frac{3}{14} x - \frac{1}{14} + \frac{9 x ^{2} - 6 x + 15}{14}) - 3 (\frac{1}{9 x ^{2} - 6 x + 15} x - \frac{1}{3} \frac{1}{9 x ^{2} - 6 x + 15} - \frac{3 x - 1}{3 9 x ^{2} - 6 x + 15} + \frac{1}{3} ln \frac{3}{14} x - \frac{1}{14} + \frac{9 x ^{2} - 6 x + 15}{14}) + C

Solution steps

\int \frac{x - 1}{3 x ^{2} - 2 x + 5} d x

Expand

\frac{x - 1}{3 x ^{2} - 2 x + 5} : \frac{x}{3 x ^{2} - 2 x + 5} - \frac{1}{3 x ^{2} - 2 x + 5}

= \int \frac{x}{3 x ^{2} - 2 x + 5} - \frac{1}{3 x ^{2} - 2 x + 5} d x

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{x}{3 x ^{2} - 2 x + 5} d x - \int \frac{1}{3 x ^{2} - 2 x + 5} d x

\int \frac{x}{3 x ^{2} - 2 x + 5} d x = \frac{1}{3 3} (9 x^{2} - 6 x + 15 + ln \frac{3}{14} x - \frac{1}{14} + \frac{9 x ^{2} - 6 x + 15}{14})

\int \frac{1}{3 x ^{2} - 2 x + 5} d x = 3 (\frac{1}{9 x ^{2} - 6 x + 15} x - \frac{1}{3} \frac{1}{9 x ^{2} - 6 x + 15} - \frac{3 x - 1}{3 9 x ^{2} - 6 x + 15} + \frac{1}{3} ln \frac{3}{14} x - \frac{1}{14} + \frac{9 x ^{2} - 6 x + 15}{14})

= \frac{1}{3 3} (9 x^{2} - 6 x + 15 + ln \frac{3}{14} x - \frac{1}{14} + \frac{9 x ^{2} - 6 x + 15}{14}) - 3 (\frac{1}{9 x ^{2} - 6 x + 15} x - \frac{1}{3} \frac{1}{9 x ^{2} - 6 x + 15} - \frac{3 x - 1}{3 9 x ^{2} - 6 x + 15} + \frac{1}{3} ln \frac{3}{14} x - \frac{1}{14} + \frac{9 x ^{2} - 6 x + 15}{14})

Add a constant to the solution

= \frac{1}{3 3} (9 x^{2} - 6 x + 15 + ln \frac{3}{14} x - \frac{1}{14} + \frac{9 x ^{2} - 6 x + 15}{14}) - 3 (\frac{1}{9 x ^{2} - 6 x + 15} x - \frac{1}{3} \frac{1}{9 x ^{2} - 6 x + 15} - \frac{3 x - 1}{3 9 x ^{2} - 6 x + 15} + \frac{1}{3} ln \frac{3}{14} x - \frac{1}{14} + \frac{9 x ^{2} - 6 x + 15}{14}) + C

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What is the integral of (x-1)/(sqrt(3x^2-2x+5)) ?
The integral of (x-1)/(sqrt(3x^2-2x+5)) is