What is the integral of 1/(2x^2+9x+4) ?

The integral of 1/(2x^2+9x+4) is -1/7 (ln| 16/7+4/7 x|-ln| 2/7+4/7 x|)+C

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integral of 1/(2x^2+9x+4)

\int \frac{1}{2 x ^{2} + 9 x + 4} d x

- \frac{1}{7} (ln \frac{16}{7} + \frac{4}{7} x - ln \frac{2}{7} + \frac{4}{7} x) + C

Solution steps

\int \frac{1}{2 x ^{2} + 9 x + 4} d x

Complete the square

2 x^{2} + 9 x + 4 : 2 (x + \frac{9}{4})^{2} - \frac{49}{8}

= \int \frac{1}{2 ( x + \frac{9}{4} ) ^{2} - \frac{49}{8}} d x

Apply u-substitution

= \int \frac{8}{16 u ^{2} - 49} 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} - 49} d u

Apply Integral Substitution

= 8 \cdot \int \frac{1}{28 ( 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}{28} \cdot \int \frac{1}{v ^{2} - 1} d v

Factor

v^{2} - 1 : - (- v^{2} + 1)

= 8 \cdot \frac{1}{28} \cdot \int \frac{1}{- ( - 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}{28} (- \int \frac{1}{- v ^{2} + 1} d v)

Use the common integral:

\int \frac{1}{- v ^{2} + 1} d v = \frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}

= 8 \cdot \frac{1}{28} (- (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}))

Substitute back

= 8 \cdot \frac{1}{28} (- (\frac{ln \frac{4}{7} ( x + \frac{9}{4} ) + 1}{2} - \frac{ln \frac{4}{7} ( x + \frac{9}{4} ) - 1}{2}))

Simplify

8 \cdot \frac{1}{28} (- (\frac{ln \frac{4}{7} ( x + \frac{9}{4} ) + 1}{2} - \frac{ln \frac{4}{7} ( x + \frac{9}{4} ) - 1}{2})) : - \frac{1}{7} (ln \frac{16}{7} + \frac{4}{7} x - ln \frac{2}{7} + \frac{4}{7} x)

= - \frac{1}{7} (ln \frac{16}{7} + \frac{4}{7} x - ln \frac{2}{7} + \frac{4}{7} x)

Add a constant to the solution

= - \frac{1}{7} (ln \frac{16}{7} + \frac{4}{7} x - ln \frac{2}{7} + \frac{4}{7} x) + C

What is the integral of 1/(2x^2+9x+4) ?
The integral of 1/(2x^2+9x+4) is -1/7 (ln| 16/7+4/7 x|-ln| 2/7+4/7 x|)+C