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

The integral of x^{-3}+sqrt(x)-3x^{1/4}+x^2 is -1/(2x^2)+2/3 x^{3/2}-12/5 x^{5/4}+(x^3)/3+C

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integral of x^{-3}+sqrt(x)-3x^{1/4}+x^2

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

- \frac{1}{2 x ^{2}} + \frac{2}{3} x^{\frac{3}{2}} - \frac{12}{5} x^{\frac{5}{4}} + \frac{x ^{3}}{3} + C

Solution steps

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

Apply exponent rule:

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

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

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

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

\int 3 x^{\frac{1}{4}} d x = \frac{12}{5} x^{\frac{5}{4}}

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

= - \frac{1}{2 x ^{2}} + \frac{2}{3} x^{\frac{3}{2}} - \frac{12}{5} x^{\frac{5}{4}} + \frac{x ^{3}}{3}

Add a constant to the solution

= - \frac{1}{2 x ^{2}} + \frac{2}{3} x^{\frac{3}{2}} - \frac{12}{5} x^{\frac{5}{4}} + \frac{x ^{3}}{3} + C

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What is the integral of x^{-3}+sqrt(x)-3x^{1/4}+x^2 ?
The integral of x^{-3}+sqrt(x)-3x^{1/4}+x^2 is -1/(2x^2)+2/3 x^{3/2}-12/5 x^{5/4}+(x^3)/3+C