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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

Expand

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

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

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

\int \frac{1}{x} d x = ln ∣ x ∣

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

Add a constant to the solution

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

\int \frac{5 x ^{3} + 7 x}{x - 1} d x

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

y = 2 x arccos (x) - 2 1 - x^{2}

n = 0 \sum \infty \frac{( n )}{n + 1}

\frac{y}{x ^{2}} \frac{d y}{d x} + e^{2 x^{3} + y^{2}} = 0

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