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# Integrate $$4x^{3} + \frac 1 x$$ with respect to $$x$$

### Question

Integrate $$4x^{3} + \frac 1 x$$ with respect to $$x$$

### Options

A) $$\ln x$$ + $$x^{4}$$ + K

B) $$x^{-1}4$$ + $$x^{4}$$ + K

C) $$12x^{2}$$ - $$x^{-2}$$ + K

D) $$\frac 1 5x^{5}$$ + $$x^{-2}$$ + K

### Explanation:

$$f(x) = 4x^3 + \cfrac{1}{x}$$
$$\int \left( 4x^3 + \cfrac{1}{x} \right)dx = \cfrac{4x^{3+1}}{3+1} + \ln x + c$$
$$=\cfrac{4x^4}{4}+\ln x + c$$
$$=x^4+\ln x + C$$

## Dicussion (1)

• $$f(x) = 4x^3 + \cfrac{1}{x}$$
$$\int \left( 4x^3 + \cfrac{1}{x} \right)dx = \cfrac{4x^{3+1}}{3+1} + \ln x + c$$
$$=\cfrac{4x^4}{4}+\ln x + c$$
$$=x^4+\ln x + C$$