What is the integral of 5/(x^3+2x^2-35x) ?

The integral of 5/(x^3+2x^2-35x) is 5(-1/35 ln|x|+1/60 ln|x-5|+1/84 ln|x+7|)+C

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integral of 5/(x^3+2x^2-35x)

\int \frac{5}{x ^{3} + 2 x ^{2} - 35 x} d x

5 (- \frac{1}{35} ln ∣ x ∣ + \frac{1}{60} ln ∣ x - 5 ∣ + \frac{1}{84} ln ∣ x + 7 ∣) + C

Solution steps

\int \frac{5}{x ^{3} + 2 x ^{2} - 35 x} d x

Take the constant out:

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

= 5 \cdot \int \frac{1}{x ^{3} + 2 x ^{2} - 35 x} d x

Take the partial fraction of

\frac{1}{x ^{3} + 2 x ^{2} - 35 x} : - \frac{1}{35 x} + \frac{1}{60 ( x - 5 )} + \frac{1}{84 ( x + 7 )}

= 5 \cdot \int - \frac{1}{35 x} + \frac{1}{60 ( x - 5 )} + \frac{1}{84 ( 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

= 5 (- \int \frac{1}{35 x} d x + \int \frac{1}{60 ( x - 5 )} d x + \int \frac{1}{84 ( x + 7 )} d x)

\int \frac{1}{35 x} d x = \frac{1}{35} ln ∣ x ∣

\int \frac{1}{60 ( x - 5 )} d x = \frac{1}{60} ln ∣ x - 5 ∣

\int \frac{1}{84 ( x + 7 )} d x = \frac{1}{84} ln ∣ x + 7 ∣

= 5 (- \frac{1}{35} ln ∣ x ∣ + \frac{1}{60} ln ∣ x - 5 ∣ + \frac{1}{84} ln ∣ x + 7 ∣)

Add a constant to the solution

= 5 (- \frac{1}{35} ln ∣ x ∣ + \frac{1}{60} ln ∣ x - 5 ∣ + \frac{1}{84} ln ∣ x + 7 ∣) + C

u^{9}

\frac{d}{d x} (2 x e^{x} + e^{x x^{2}})

\int 2 cos (9 x) d x

y = 4 - x^{2}, x = 0, x = 2

\frac{d}{d x} (tan (x) - 0.5 x)

What is the integral of 5/(x^3+2x^2-35x) ?
The integral of 5/(x^3+2x^2-35x) is 5(-1/35 ln|x|+1/60 ln|x-5|+1/84 ln|x+7|)+C