What is the integral of (4x^2+x^{-3}) ?

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

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integral of (4x^2+x^{-3})

\int (4 x^{2} + x^{- 3}) d x

\frac{4 x ^{3}}{3} - \frac{1}{2 x ^{2}} + C

Solution steps

\int 4 x^{2} + x^{- 3} d x

Apply exponent rule:

a^{- b} = \frac{1}{a ^{b}}

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

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

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

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

Add a constant to the solution

= \frac{4 x ^{3}}{3} - \frac{1}{2 x ^{2}} + C

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