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

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

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integral of (4x^2-1)/x

\int \frac{4 x ^{2} - 1}{x} d x

2 x^{2} - ln ∣ x ∣ + C

Solution steps

\int \frac{4 x ^{2} - 1}{x} d x

Expand

\frac{4 x ^{2} - 1}{x} : 4 x - \frac{1}{x}

= \int 4 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

= \int 4 x d x - \int \frac{1}{x} d x

\int 4 x d x = 2 x^{2}

\int \frac{1}{x} d x = ln ∣ x ∣

= 2 x^{2} - ln ∣ x ∣

Add a constant to the solution

= 2 x^{2} - ln ∣ x ∣ + C

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