What is the integral of 1/(2x^2-3x+9) ?

The integral of 1/(2x^2-3x+9) is 2/(3sqrt(7))arctan(4/(3sqrt(7))x-1/(sqrt(7)))+C

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integral of 1/(2x^2-3x+9)

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

\frac{2}{3 7} arctan (\frac{4}{3 7} x - \frac{1}{7}) + C

Solution steps

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

Complete the square

2 x^{2} - 3 x + 9 : 2 (x - \frac{3}{4})^{2} + \frac{63}{8}

= \int \frac{1}{2 ( x - \frac{3}{4} ) ^{2} + \frac{63}{8}} d x

Apply u-substitution

: \int \frac{8}{16 u ^{2} + 63} d u

= \int \frac{8}{16 u ^{2} + 63} d u

Take the constant out:

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

= 8 \cdot \int \frac{1}{16 u ^{2} + 63} d u

Apply Integral Substitution

= 8 \cdot \int \frac{1}{12 7 ( v ^{2} + 1 )} d v

Take the constant out:

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

= 8 \cdot \frac{1}{12 7} \cdot \int \frac{1}{v ^{2} + 1} d v

Use the common integral:

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

= 8 \cdot \frac{1}{12 7} arctan (v)

Substitute back

= 8 \cdot \frac{1}{12 7} arctan (\frac{4}{3 7} (x - \frac{3}{4}))

Simplify

8 \cdot \frac{1}{12 7} arctan (\frac{4}{3 7} (x - \frac{3}{4})) : \frac{2}{3 7} arctan (\frac{4}{3 7} x - \frac{1}{7})

= \frac{2}{3 7} arctan (\frac{4}{3 7} x - \frac{1}{7})

Add a constant to the solution

= \frac{2}{3 7} arctan (\frac{4}{3 7} x - \frac{1}{7}) + C

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What is the integral of 1/(2x^2-3x+9) ?
The integral of 1/(2x^2-3x+9) is 2/(3sqrt(7))arctan(4/(3sqrt(7))x-1/(sqrt(7)))+C