What is the integral of 1/(-x^2+249x+250) ?

The integral of 1/(-x^2+249x+250) is 1/251 ln|x+1|-1/251 ln|x-250|+C

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integral of 1/(-x^2+249x+250)

\int \frac{1}{- x ^{2} + 249 x + 250} d x

\frac{1}{251} ln ∣ x + 1 ∣ - \frac{1}{251} ln ∣ x - 250 ∣ + C

Solution steps

\int \frac{1}{- x ^{2} + 249 x + 250} d x

Take the partial fraction of

\frac{1}{- x ^{2} + 249 x + 250} : \frac{1}{251 ( x + 1 )} - \frac{1}{251 ( x - 250 )}

= \int \frac{1}{251 ( x + 1 )} - \frac{1}{251 ( x - 250 )} 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{1}{251 ( x + 1 )} d x - \int \frac{1}{251 ( x - 250 )} d x

\int \frac{1}{251 ( x + 1 )} d x = \frac{1}{251} ln ∣ x + 1 ∣

\int \frac{1}{251 ( x - 250 )} d x = \frac{1}{251} ln ∣ x - 250 ∣

= \frac{1}{251} ln ∣ x + 1 ∣ - \frac{1}{251} ln ∣ x - 250 ∣

Add a constant to the solution

= \frac{1}{251} ln ∣ x + 1 ∣ - \frac{1}{251} ln ∣ x - 250 ∣ + C

x \to \infty lim (\frac{( ln ( 3 x ) )}{e ^{7 x}})

n = 0 \sum \infty (- 5)^{n}

\int_{0}^{1} (1 + 4 x y) d x

y = 3 + 4 x^{2} - 2 x^{3}, (2, 3)

\frac{d}{d x} (\frac{4 + csc ( x )}{8 - csc ( x )})

What is the integral of 1/(-x^2+249x+250) ?
The integral of 1/(-x^2+249x+250) is 1/251 ln|x+1|-1/251 ln|x-250|+C