What is the integral of (196)/(x^3-14x^2) ?

The integral of (196)/(x^3-14x^2) is -ln|x|+(14)/x+ln|x-14|+C

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integral of (196)/(x^3-14x^2)

\int \frac{196}{x ^{3} - 14 x ^{2}} d x

- ln ∣ x ∣ + \frac{14}{x} + ln ∣ x - 14 ∣ + C

Solution steps

\int \frac{196}{x ^{3} - 14 x ^{2}} d x

Take the constant out:

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

= 196 \cdot \int \frac{1}{x ^{3} - 14 x ^{2}} d x

Take the partial fraction of

\frac{1}{x ^{3} - 14 x ^{2}} : - \frac{1}{196 x} - \frac{1}{14 x ^{2}} + \frac{1}{196 ( x - 14 )}

= 196 \cdot \int - \frac{1}{196 x} - \frac{1}{14 x ^{2}} + \frac{1}{196 ( x - 14 )} 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

= 196 (- \int \frac{1}{196 x} d x - \int \frac{1}{14 x ^{2}} d x + \int \frac{1}{196 ( x - 14 )} d x)

\int \frac{1}{196 x} d x = \frac{1}{196} ln ∣ x ∣

\int \frac{1}{14 x ^{2}} d x = - \frac{1}{14 x}

\int \frac{1}{196 ( x - 14 )} d x = \frac{1}{196} ln ∣ x - 14 ∣

= 196 (- \frac{1}{196} ln ∣ x ∣ - (- \frac{1}{14 x}) + \frac{1}{196} ln ∣ x - 14 ∣)

Simplify

196 (- \frac{1}{196} ln ∣ x ∣ - (- \frac{1}{14 x}) + \frac{1}{196} ln ∣ x - 14 ∣) : - ln ∣ x ∣ + \frac{14}{x} + ln ∣ x - 14 ∣

= - ln ∣ x ∣ + \frac{14}{x} + ln ∣ x - 14 ∣

Add a constant to the solution

= - ln ∣ x ∣ + \frac{14}{x} + ln ∣ x - 14 ∣ + C

\int \frac{1}{x 324 x ^{2} - 324} d x

\int \frac{9 - 162 x}{16 - 81 x ^{2}} d x

\int_{0}^{1} x cos (5 π x) d x

y = (3 x - 1)^{2}

3 x^{2} + 3

What is the integral of (196)/(x^3-14x^2) ?
The integral of (196)/(x^3-14x^2) is -ln|x|+(14)/x+ln|x-14|+C