What is the integral of ((6x^2-3x)/(4x^3)) ?

The integral of ((6x^2-3x)/(4x^3)) is 1/4 (6ln|x|+3/x)+C

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integral of ((6x^2-3x)/(4x^3))

\int (\frac{6 x ^{2} - 3 x}{4 x ^{3}}) d x

\frac{1}{4} (6 ln ∣ x ∣ + \frac{3}{x}) + C

Solution steps

\int \frac{6 x ^{2} - 3 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{6 x ^{2} - 3 x}{x ^{3}} d x

Expand

\frac{6 x ^{2} - 3 x}{x ^{3}} : \frac{6}{x} - \frac{3}{x ^{2}}

= \frac{1}{4} \cdot \int \frac{6}{x} - \frac{3}{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 \frac{6}{x} d x - \int \frac{3}{x ^{2}} d x)

\int \frac{6}{x} d x = 6 ln ∣ x ∣

\int \frac{3}{x ^{2}} d x = - \frac{3}{x}

= \frac{1}{4} (6 ln ∣ x ∣ - (- \frac{3}{x}))

Simplify

= \frac{1}{4} (6 ln ∣ x ∣ + \frac{3}{x})

Add a constant to the solution

= \frac{1}{4} (6 ln ∣ x ∣ + \frac{3}{x}) + C

2 (8^{x})

y^{^{''}} + 21 y = 0

\int x^{2} (\frac{x ^{3}}{12} + 4)^{3} d x

x \to \infty lim ((1 + \frac{n + 3}{4 n + 1})^{2 n})

\int θ (θ - 2)^{3} d θ

What is the integral of ((6x^2-3x)/(4x^3)) ?
The integral of ((6x^2-3x)/(4x^3)) is 1/4 (6ln|x|+3/x)+C