What is the integral of 1/(45-20x-25x^2) ?

The integral of 1/(45-20x-25x^2) is 1/70 (ln| 9/7+5/7 x|-ln|-5/7+5/7 x|)+C

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integral of 1/(45-20x-25x^2)

\int \frac{1}{45 - 20 x - 25 x ^{2}} d x

\frac{1}{70} (ln \frac{9}{7} + \frac{5}{7} x - ln - \frac{5}{7} + \frac{5}{7} x) + C

Solution steps

\int \frac{1}{45 - 20 x - 25 x ^{2}} d x

Complete the square

45 - 20 x - 25 x^{2} : - 25 (x + \frac{2}{5})^{2} + 49

= \int \frac{1}{- 25 ( x + \frac{2}{5} ) ^{2} + 49} d x

Apply u-substitution

: \int \frac{1}{- 25 u ^{2} + 49} d u

= \int \frac{1}{- 25 u ^{2} + 49} d u

Apply Integral Substitution

= \int \frac{1}{35 ( - v ^{2} + 1 )} d v

Take the constant out:

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

= \frac{1}{35} \cdot \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}

= \frac{1}{35} (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2})

Substitute back

= \frac{1}{35} (\frac{ln \frac{5}{7} ( x + \frac{2}{5} ) + 1}{2} - \frac{ln \frac{5}{7} ( x + \frac{2}{5} ) - 1}{2})

Simplify

\frac{1}{35} (\frac{ln \frac{5}{7} ( x + \frac{2}{5} ) + 1}{2} - \frac{ln \frac{5}{7} ( x + \frac{2}{5} ) - 1}{2}) : \frac{1}{70} (ln \frac{9}{7} + \frac{5}{7} x - ln - \frac{5}{7} + \frac{5}{7} x)

= \frac{1}{70} (ln \frac{9}{7} + \frac{5}{7} x - ln - \frac{5}{7} + \frac{5}{7} x)

Add a constant to the solution

= \frac{1}{70} (ln \frac{9}{7} + \frac{5}{7} x - ln - \frac{5}{7} + \frac{5}{7} x) + C

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What is the integral of 1/(45-20x-25x^2) ?
The integral of 1/(45-20x-25x^2) is 1/70 (ln| 9/7+5/7 x|-ln|-5/7+5/7 x|)+C