What is the integral of 2/9 (x^2)/(x^5) ?

The integral of 2/9 (x^2)/(x^5) is -1/(9x^2)+C

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integral of 2/9 (x^2)/(x^5)

\int \frac{2}{9} \frac{x ^{2}}{x ^{5}} d x

- \frac{1}{9 x ^{2}} + C

Solution steps

\int \frac{2}{9} \cdot \frac{x ^{2}}{x ^{5}} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{2}{9} \cdot \int \frac{x ^{2}}{x ^{5}} d x

Simplify

\frac{x ^{2}}{x ^{5}} : \frac{1}{x ^{3}}

= \frac{2}{9} \cdot \int \frac{1}{x ^{3}} d x

Apply the Power Rule

: - \frac{1}{2 x ^{2}}

= \frac{2}{9} (- \frac{1}{2 x ^{2}})

Simplify

\frac{2}{9} (- \frac{1}{2 x ^{2}}) : - \frac{1}{9 x ^{2}}

= - \frac{1}{9 x ^{2}}

Add a constant to the solution

= - \frac{1}{9 x ^{2}} + C

\int \frac{1}{( x + 1 ) ( x + 2 ) ( x + 3 )} d x

\frac{\partial}{\partial x} (9 - x^{2} + x y - 4 y^{2})

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

x \to 0 lim (\frac{5 ^{x} - 1}{x})

y = x^{e}

What is the integral of 2/9 (x^2)/(x^5) ?
The integral of 2/9 (x^2)/(x^5) is -1/(9x^2)+C