What is the integral of 1/(x^2)+20x^{1/4} ?

The integral of 1/(x^2)+20x^{1/4} is -1/x+16x^{5/4}+C

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integral of 1/(x^2)+20x^{1/4}

\int \frac{1}{x ^{2}} + 20 x^{\frac{1}{4}} d x

- \frac{1}{x} + 16 x^{\frac{5}{4}} + C

Solution steps

\int \frac{1}{x ^{2}} + 20 x^{\frac{1}{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 \frac{1}{x ^{2}} d x + \int 20 x^{\frac{1}{4}} d x

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

\int 20 x^{\frac{1}{4}} d x = 16 x^{\frac{5}{4}}

= - \frac{1}{x} + 16 x^{\frac{5}{4}}

Add a constant to the solution

= - \frac{1}{x} + 16 x^{\frac{5}{4}} + C

x \to 2 lim (x - 2)

\int \frac{1}{x \cdot ( 1 - x )} d x

f (x) = x + 5

sin (2 x), π

\int_{0}^{π} x \cdot sin (2 x) d x

What is the integral of 1/(x^2)+20x^{1/4} ?
The integral of 1/(x^2)+20x^{1/4} is -1/x+16x^{5/4}+C