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

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

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integral of (6+x^2)/x

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

6 ln ∣ x ∣ + \frac{x ^{2}}{2} + C

Solution steps

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

Expand

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

= \int \frac{6}{x} + 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 \frac{6}{x} d x + \int x d x

\int \frac{6}{x} d x = 6 ln ∣ x ∣

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

= 6 ln ∣ x ∣ + \frac{x ^{2}}{2}

Add a constant to the solution

= 6 ln ∣ x ∣ + \frac{x ^{2}}{2} + C

(x + 1) (7 x + 2)

\frac{\partial}{\partial x} (\frac{x}{y ^{2}} + x y + 8)

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

z (z + 1) e^{z}

f (x) = 16 - x^{2}, (1, 4)

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