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

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

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

\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}) + C

Solution steps

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

Complete the square

3 x^{2} - 2 x + 5 : 3 (x - \frac{1}{3})^{2} + \frac{14}{3}

= \int \frac{1}{3 ( x - \frac{1}{3} ) ^{2} + \frac{14}{3}} d x

Apply u-substitution

: \int \frac{3}{9 u ^{2} + 14} d u

= \int \frac{3}{9 u ^{2} + 14} d u

\frac{3}{9 u ^{2} + 14} = 3 \frac{1}{9 u ^{2} + 14}

= \int 3 \frac{1}{9 u ^{2} + 14} d u

Take the constant out:

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

= 3 \cdot \int \frac{1}{9 u ^{2} + 14} d u

Apply Integration By Parts

: \frac{1}{9 u ^{2} + 14} u - \int - \frac{9 u ^{2}}{( 9 u ^{2} + 14 ) ^{\frac{3}{2}}} d u

= 3 (\frac{1}{9 u ^{2} + 14} u - \int - \frac{9 u ^{2}}{( 9 u ^{2} + 14 ) ^{\frac{3}{2}}} d u)

\int - \frac{9 u ^{2}}{( 9 u ^{2} + 14 ) ^{\frac{3}{2}}} d u = \frac{u}{14 + 9 u ^{2}} - \frac{1}{3} ln \frac{3}{14} u + \frac{1}{14} (14 + 9 u^{2})

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

Substitute back

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

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

Simplify

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

= 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

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