What is the integral of ((3x^2+4x+1))/(2x) ?

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

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integral of ((3x^2+4x+1))/(2x)

\int \frac{( 3 x ^{2} + 4 x + 1 )}{2 x} d x

\frac{1}{2} (\frac{3 x ^{2}}{2} + 4 x + ln ∣ x ∣) + C

Solution steps

\int \frac{3 x ^{2} + 4 x + 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} + 4 x + 1}{x} d x

Expand

\frac{3 x ^{2} + 4 x + 1}{x} : 3 x + 4 + \frac{1}{x}

= \frac{1}{2} \cdot \int 3 x + 4 + \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 4 d x + \int \frac{1}{x} d x)

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

\int 4 d x = 4 x

\int \frac{1}{x} d x = ln ∣ x ∣

= \frac{1}{2} (\frac{3 x ^{2}}{2} + 4 x + ln ∣ x ∣)

Add a constant to the solution

= \frac{1}{2} (\frac{3 x ^{2}}{2} + 4 x + ln ∣ x ∣) + C

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