What is the integral of 1/(16x^2-24x-7) ?

The integral of 1/(16x^2-24x-7) is -1/32 (ln|x+1/4 |-ln|x-7/4 |)+C

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integral of 1/(16x^2-24x-7)

\int \frac{1}{16 x ^{2} - 24 x - 7} d x

- \frac{1}{32} (ln x + \frac{1}{4} - ln x - \frac{7}{4}) + C

Solution steps

\int \frac{1}{16 x ^{2} - 24 x - 7} d x

Complete the square

16 x^{2} - 24 x - 7 : 16 (x - \frac{3}{4})^{2} - 16

= \int \frac{1}{16 ( x - \frac{3}{4} ) ^{2} - 16} d x

Apply u-substitution

: \int \frac{1}{16 u ^{2} - 16} d u

= \int \frac{1}{16 u ^{2} - 16} d u

\frac{1}{16 u ^{2} - 16} = \frac{1}{16} \frac{1}{u ^{2} - 1}

= \int \frac{1}{16} \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}{16} \cdot \int \frac{1}{u ^{2} - 1} d u

Factor

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

= \frac{1}{16} \cdot \int \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}{16} (- \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}{16} (- (\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}))

Substitute back

u = x - \frac{3}{4}

= \frac{1}{16} (- (\frac{ln x - \frac{3}{4} + 1}{2} - \frac{ln x - \frac{3}{4} - 1}{2}))

Simplify

\frac{1}{16} (- (\frac{ln x - \frac{3}{4} + 1}{2} - \frac{ln x - \frac{3}{4} - 1}{2})) : - \frac{1}{32} (ln x + \frac{1}{4} - ln x - \frac{7}{4})

= - \frac{1}{32} (ln x + \frac{1}{4} - ln x - \frac{7}{4})

Add a constant to the solution

= - \frac{1}{32} (ln x + \frac{1}{4} - ln x - \frac{7}{4}) + C

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What is the integral of 1/(16x^2-24x-7) ?
The integral of 1/(16x^2-24x-7) is -1/32 (ln|x+1/4 |-ln|x-7/4 |)+C