What is the integral of (3x-1)/(x^{2/3)} ?

The integral of (3x-1)/(x^{2/3)} is 9/4 x^{4/3}-3x^{1/3}+C

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integral of (3x-1)/(x^{2/3)}

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

\frac{9}{4} x^{\frac{4}{3}} - 3 x^{\frac{1}{3}} + C

Solution steps

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

Expand

\frac{3 x - 1}{x ^{\frac{2}{3}}} : 3 x^{\frac{1}{3}} - \frac{1}{x ^{\frac{2}{3}}}

= \int 3 x^{\frac{1}{3}} - \frac{1}{x ^{\frac{2}{3}}} 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

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

\int 3 x^{\frac{1}{3}} d x = \frac{9}{4} x^{\frac{4}{3}}

\int \frac{1}{x ^{\frac{2}{3}}} d x = 3 x^{\frac{1}{3}}

= \frac{9}{4} x^{\frac{4}{3}} - 3 x^{\frac{1}{3}}

Add a constant to the solution

= \frac{9}{4} x^{\frac{4}{3}} - 3 x^{\frac{1}{3}} + C

\int \frac{( x - 1 )}{x ^{3} + x ^{2} + 6 x} d x

x \to 4 - lim (4 x)

\frac{d}{d x} (\frac{1}{3 x ^{2} + 1})

\int x^{2} e^{4 x^{3}} d x

n = 0 \sum \infty 1^{l n (n)}

What is the integral of (3x-1)/(x^{2/3)} ?
The integral of (3x-1)/(x^{2/3)} is 9/4 x^{4/3}-3x^{1/3}+C