What is the integral of (49)/(x^3-7x^2) ?

The integral of (49)/(x^3-7x^2) is -ln|x|+7/x+ln|x-7|+C

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integral of (49)/(x^3-7x^2)

\int \frac{49}{x ^{3} - 7 x ^{2}} d x

- ln ∣ x ∣ + \frac{7}{x} + ln ∣ x - 7 ∣ + C

Solution steps

\int \frac{49}{x ^{3} - 7 x ^{2}} d x

Take the constant out:

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

= 49 \cdot \int \frac{1}{x ^{3} - 7 x ^{2}} d x

Take the partial fraction of

\frac{1}{x ^{3} - 7 x ^{2}} : - \frac{1}{49 x} - \frac{1}{7 x ^{2}} + \frac{1}{49 ( x - 7 )}

= 49 \cdot \int - \frac{1}{49 x} - \frac{1}{7 x ^{2}} + \frac{1}{49 ( 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

= 49 (- \int \frac{1}{49 x} d x - \int \frac{1}{7 x ^{2}} d x + \int \frac{1}{49 ( x - 7 )} d x)

\int \frac{1}{49 x} d x = \frac{1}{49} ln ∣ x ∣

\int \frac{1}{7 x ^{2}} d x = - \frac{1}{7 x}

\int \frac{1}{49 ( x - 7 )} d x = \frac{1}{49} ln ∣ x - 7 ∣

= 49 (- \frac{1}{49} ln ∣ x ∣ - (- \frac{1}{7 x}) + \frac{1}{49} ln ∣ x - 7 ∣)

Simplify

49 (- \frac{1}{49} ln ∣ x ∣ - (- \frac{1}{7 x}) + \frac{1}{49} ln ∣ x - 7 ∣) : - ln ∣ x ∣ + \frac{7}{x} + ln ∣ x - 7 ∣

= - ln ∣ x ∣ + \frac{7}{x} + ln ∣ x - 7 ∣

Add a constant to the solution

= - ln ∣ x ∣ + \frac{7}{x} + ln ∣ x - 7 ∣ + C

\frac{d y}{d x} + 12 y = 5

\frac{d}{d x} ((cos (x)) x)

x^{2} \frac{d y}{d x} = y - x y, y (- 1) = - 2

f (x) = ln (x^{3} + 10)

\int x^{\frac{7}{5}} d x

What is the integral of (49)/(x^3-7x^2) ?
The integral of (49)/(x^3-7x^2) is -ln|x|+7/x+ln|x-7|+C