What is the integral of 1/(35+12x-36x^2) ?

The integral of 1/(35+12x-36x^2) is 1/72 (ln|x+5/6 |-ln|x-7/6 |)+C

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integral of 1/(35+12x-36x^2)

\int \frac{1}{35 + 12 x - 36 x ^{2}} d x

\frac{1}{72} (ln x + \frac{5}{6} - ln x - \frac{7}{6}) + C

Solution steps

\int \frac{1}{35 + 12 x - 36 x ^{2}} d x

Complete the square

35 + 12 x - 36 x^{2} : - 36 (x - \frac{1}{6})^{2} + 36

= \int \frac{1}{- 36 ( x - \frac{1}{6} ) ^{2} + 36} d x

Apply u-substitution

: \int \frac{1}{- 36 u ^{2} + 36} d u

= \int \frac{1}{- 36 u ^{2} + 36} d u

\frac{1}{- 36 u ^{2} + 36} = \frac{1}{36} \frac{1}{- u ^{2} + 1}

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

Take the constant out:

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

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

Use the common integral:

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

= \frac{1}{36} (\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2})

Substitute back

u = x - \frac{1}{6}

= \frac{1}{36} (\frac{ln x - \frac{1}{6} + 1}{2} - \frac{ln x - \frac{1}{6} - 1}{2})

Simplify

\frac{1}{36} (\frac{ln x - \frac{1}{6} + 1}{2} - \frac{ln x - \frac{1}{6} - 1}{2}) : \frac{1}{72} (ln x + \frac{5}{6} - ln x - \frac{7}{6})

= \frac{1}{72} (ln x + \frac{5}{6} - ln x - \frac{7}{6})

Add a constant to the solution

= \frac{1}{72} (ln x + \frac{5}{6} - ln x - \frac{7}{6}) + C

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What is the integral of 1/(35+12x-36x^2) ?
The integral of 1/(35+12x-36x^2) is 1/72 (ln|x+5/6 |-ln|x-7/6 |)+C