What is the integral of (13x^{14}-5x^6+2x)/(4x^3) ?

The integral of (13x^{14}-5x^6+2x)/(4x^3) is 1/4 ((13x^{12}}{12}-\frac{5x^4)/4-2/x)+C

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integral of (13x^{14}-5x^6+2x)/(4x^3)

\int \frac{13 x ^{14} - 5 x ^{6} + 2 x}{4 x ^{3}} d x

\frac{1}{4} (\frac{13 x ^{12}}{12} - \frac{5 x ^{4}}{4} - \frac{2}{x}) + C

Solution steps

\int \frac{13 x ^{14} - 5 x ^{6} + 2 x}{4 x ^{3}} d x

Take the constant out:

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

= \frac{1}{4} \cdot \int \frac{13 x ^{14} - 5 x ^{6} + 2 x}{x ^{3}} d x

Expand

\frac{13 x ^{14} - 5 x ^{6} + 2 x}{x ^{3}} : 13 x^{11} - 5 x^{3} + \frac{2}{x ^{2}}

= \frac{1}{4} \cdot \int 13 x^{11} - 5 x^{3} + \frac{2}{x ^{2}} 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

= \frac{1}{4} (\int 13 x^{11} d x - \int 5 x^{3} d x + \int \frac{2}{x ^{2}} d x)

\int 13 x^{11} d x = \frac{13 x ^{12}}{12}

\int 5 x^{3} d x = \frac{5 x ^{4}}{4}

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

= \frac{1}{4} (\frac{13 x ^{12}}{12} - \frac{5 x ^{4}}{4} - \frac{2}{x})

Add a constant to the solution

= \frac{1}{4} (\frac{13 x ^{12}}{12} - \frac{5 x ^{4}}{4} - \frac{2}{x}) + C

What is the integral of (13x^{14}-5x^6+2x)/(4x^3) ?
The integral of (13x^{14}-5x^6+2x)/(4x^3) is 1/4 ((13x^{12}}{12}-\frac{5x^4)/4-2/x)+C