What is the integral of 1/(-2x^2+13x-15) ?

The integral of 1/(-2x^2+13x-15) is 1/7 ln|2x-3|-1/7 ln|x-5|+C

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integral of 1/(-2x^2+13x-15)

\int \frac{1}{- 2 x ^{2} + 13 x - 15} d x

\frac{1}{7} ln ∣ 2 x - 3 ∣ - \frac{1}{7} ln ∣ x - 5 ∣ + C

Solution steps

\int \frac{1}{- 2 x ^{2} + 13 x - 15} d x

Factor

\frac{1}{- 2 x ^{2} + 13 x - 15}

= \int \frac{1}{- ( 2 x - 3 ) ( x - 5 )} d x

Take the constant out:

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

= - \int \frac{1}{( 2 x - 3 ) ( x - 5 )} d x

Take the partial fraction of

\frac{1}{( 2 x - 3 ) ( x - 5 )} : - \frac{2}{7 ( 2 x - 3 )} + \frac{1}{7 ( x - 5 )}

= - \int - \frac{2}{7 ( 2 x - 3 )} + \frac{1}{7 ( x - 5 )} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - (- \int \frac{2}{7 ( 2 x - 3 )} d x + \int \frac{1}{7 ( x - 5 )} d x)

\int \frac{2}{7 ( 2 x - 3 )} d x = \frac{1}{7} ln ∣ 2 x - 3 ∣

\int \frac{1}{7 ( x - 5 )} d x = \frac{1}{7} ln ∣ x - 5 ∣

= - (- \frac{1}{7} ln ∣ 2 x - 3 ∣ + \frac{1}{7} ln ∣ x - 5 ∣)

Simplify

= \frac{1}{7} ln ∣ 2 x - 3 ∣ - \frac{1}{7} ln ∣ x - 5 ∣

Add a constant to the solution

= \frac{1}{7} ln ∣ 2 x - 3 ∣ - \frac{1}{7} ln ∣ x - 5 ∣ + C

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What is the integral of 1/(-2x^2+13x-15) ?
The integral of 1/(-2x^2+13x-15) is 1/7 ln|2x-3|-1/7 ln|x-5|+C