What is the integral of 4/((x^2+5x-14)) ?

The integral of 4/((x^2+5x-14)) is 4(1/9 ln|x-2|-1/9 ln|x+7|)+C

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integral of 4/((x^2+5x-14))

\int \frac{4}{( x ^{2} + 5 x - 14 )} d x

4 (\frac{1}{9} ln ∣ x - 2 ∣ - \frac{1}{9} ln ∣ x + 7 ∣) + C

Solution steps

\int \frac{4}{x ^{2} + 5 x - 14} d x

Take the constant out:

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

= 4 \cdot \int \frac{1}{x ^{2} + 5 x - 14} d x

Take the partial fraction of

\frac{1}{x ^{2} + 5 x - 14} : \frac{1}{9 ( x - 2 )} - \frac{1}{9 ( x + 7 )}

= 4 \cdot \int \frac{1}{9 ( x - 2 )} - \frac{1}{9 ( x + 7 )} 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

= 4 (\int \frac{1}{9 ( x - 2 )} d x - \int \frac{1}{9 ( x + 7 )} d x)

\int \frac{1}{9 ( x - 2 )} d x = \frac{1}{9} ln ∣ x - 2 ∣

\int \frac{1}{9 ( x + 7 )} d x = \frac{1}{9} ln ∣ x + 7 ∣

= 4 (\frac{1}{9} ln ∣ x - 2 ∣ - \frac{1}{9} ln ∣ x + 7 ∣)

Add a constant to the solution

= 4 (\frac{1}{9} ln ∣ x - 2 ∣ - \frac{1}{9} ln ∣ x + 7 ∣) + C

x \to 2 lim (2 x^{2} - 3)

x \to - 1 + lim (\frac{1}{x ^{2} - 1})

\int \frac{2 x ^{3} + x ^{2} - 21 x + 24}{x ^{2} + 2 x - 8} d x

\int 9 x^{2} - 4 d x

f (x) = 4 x^{2} + 2, at x = - 2

What is the integral of 4/((x^2+5x-14)) ?
The integral of 4/((x^2+5x-14)) is 4(1/9 ln|x-2|-1/9 ln|x+7|)+C