What is the integral of (x^2-5x+1)/(3x^2) ?

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

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integral of (x^2-5x+1)/(3x^2)

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

\frac{1}{3} (x - 5 ln ∣ x ∣ - \frac{1}{x}) + C

Solution steps

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

Take the constant out:

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

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

Expand

\frac{x ^{2} - 5 x + 1}{x ^{2}} : 1 - \frac{5}{x} + \frac{1}{x ^{2}}

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

\int 1 d x = x

\int \frac{5}{x} d x = 5 ln ∣ x ∣

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

= \frac{1}{3} (x - 5 ln ∣ x ∣ - \frac{1}{x})

Add a constant to the solution

= \frac{1}{3} (x - 5 ln ∣ x ∣ - \frac{1}{x}) + C

\frac{\partial}{\partial x} (2 sin (5 x) cos (4 y))

\int sin^{2} (t) cos^{2} (t) d t

\frac{d}{d x} (x^{6 c o s (x)})

\frac{2}{1 + x ^{2}}

y^{^{'}} + y = e^{3 x}

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