What is the integral of 1/(3x^2-5x+4) ?

The integral of 1/(3x^2-5x+4) is 2/(sqrt(23))arctan((6x-5)/(sqrt(23)))+C

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integral of 1/(3x^2-5x+4)

\int \frac{1}{3 x ^{2} - 5 x + 4} d x

\frac{2}{23} arctan (\frac{6 x - 5}{23}) + C

Solution steps

\int \frac{1}{3 x ^{2} - 5 x + 4} d x

Complete the square

3 x^{2} - 5 x + 4 : 3 (x - \frac{5}{6})^{2} + \frac{23}{12}

= \int \frac{1}{3 ( x - \frac{5}{6} ) ^{2} + \frac{23}{12}} d x

Apply u-substitution

: \int \frac{12}{36 u ^{2} + 23} d u

= \int \frac{12}{36 u ^{2} + 23} d u

Take the constant out:

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

= 12 \cdot \int \frac{1}{36 u ^{2} + 23} d u

Apply Integral Substitution

= 12 \cdot \int \frac{1}{6 23 ( v ^{2} + 1 )} d v

Take the constant out:

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

= 12 \cdot \frac{1}{6 23} \cdot \int \frac{1}{v ^{2} + 1} d v

Use the common integral:

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

= 12 \cdot \frac{1}{6 23} arctan (v)

Substitute back

= 12 \cdot \frac{1}{6 23} arctan (\frac{6}{23} (x - \frac{5}{6}))

Simplify

12 \cdot \frac{1}{6 23} arctan (\frac{6}{23} (x - \frac{5}{6})) : \frac{2}{23} arctan (\frac{6 x - 5}{23})

= \frac{2}{23} arctan (\frac{6 x - 5}{23})

Add a constant to the solution

= \frac{2}{23} arctan (\frac{6 x - 5}{23}) + C

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What is the integral of 1/(3x^2-5x+4) ?
The integral of 1/(3x^2-5x+4) is 2/(sqrt(23))arctan((6x-5)/(sqrt(23)))+C