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

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

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

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

\frac{1}{3} (x^{2} + \frac{5}{x}) + C

Solution steps

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

Expand

\frac{2 x ^{3} - 5}{x ^{2}} : 2 x - \frac{5}{x ^{2}}

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

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

\int \frac{5}{x ^{2}} d x = - \frac{5}{x}

= \frac{1}{3} (x^{2} - (- \frac{5}{x}))

Simplify

= \frac{1}{3} (x^{2} + \frac{5}{x})

Add a constant to the solution

= \frac{1}{3} (x^{2} + \frac{5}{x}) + C

\int \frac{1}{7 x ( 1 + 7 x )} d x

\frac{d}{d x} (\frac{1}{g ( x ) ( x ) ^{2} ( x )})

\int \frac{( 4 x - 3 )}{x + 1} d x

x \to 0 lim (- (e^{\frac{1}{x}}))

f (x) = 2 x^{2} + 3

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