What is the integral of (6x^2+1)/(8x) ?

The integral of (6x^2+1)/(8x) is 1/8 (3x^2+ln|x|)+C

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integral of (6x^2+1)/(8x)

\int \frac{6 x ^{2} + 1}{8 x} d x

\frac{1}{8} (3 x^{2} + ln ∣ x ∣) + C

Solution steps

\int \frac{6 x ^{2} + 1}{8 x} d x

Take the constant out:

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

= \frac{1}{8} \cdot \int \frac{6 x ^{2} + 1}{x} d x

Expand

\frac{6 x ^{2} + 1}{x} : 6 x + \frac{1}{x}

= \frac{1}{8} \cdot \int 6 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}{8} (\int 6 x d x + \int \frac{1}{x} d x)

\int 6 x d x = 3 x^{2}

\int \frac{1}{x} d x = ln ∣ x ∣

= \frac{1}{8} (3 x^{2} + ln ∣ x ∣)

Add a constant to the solution

= \frac{1}{8} (3 x^{2} + ln ∣ x ∣) + C

What is the integral of (6x^2+1)/(8x) ?
The integral of (6x^2+1)/(8x) is 1/8 (3x^2+ln|x|)+C