What is the integral of 6x^2-3/(2x^4) ?

The integral of 6x^2-3/(2x^4) is 2x^3+1/(2x^3)+C

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integral of 6x^2-3/(2x^4)

\int 6 x^{2} - \frac{3}{2 x ^{4}} d x

2 x^{3} + \frac{1}{2 x ^{3}} + C

Solution steps

\int 6 x^{2} - \frac{3}{2 x ^{4}} 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 6 x^{2} d x - \int \frac{3}{2 x ^{4}} d x

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

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

= 2 x^{3} - (- \frac{1}{2 x ^{3}})

Simplify

= 2 x^{3} + \frac{1}{2 x ^{3}}

Add a constant to the solution

= 2 x^{3} + \frac{1}{2 x ^{3}} + C

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What is the integral of 6x^2-3/(2x^4) ?
The integral of 6x^2-3/(2x^4) is 2x^3+1/(2x^3)+C