What is the integral of 1/(aauu) ?

The integral of 1/(aauu) is -1/(a^2u)+C

Popular >

integral of 1/(aauu)

\int \frac{1}{aa uu} d u

- \frac{1}{a ^{2} u} + C

Solution steps

\int \frac{1}{aa uu} d u

Take the constant out:

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

= \frac{1}{aa} \cdot \int \frac{1}{uu} d u

aa = a^{2}

= \frac{1}{a ^{2}} \cdot \int \frac{1}{uu} d u

uu = u^{2}

= \frac{1}{a ^{2}} \cdot \int \frac{1}{u ^{2}} d u

Apply the Power Rule

: - \frac{1}{u}

= \frac{1}{a ^{2}} (- \frac{1}{u})

Simplify

\frac{1}{a ^{2}} (- \frac{1}{u}) : - \frac{1}{a ^{2} u}

= - \frac{1}{a ^{2} u}

Add a constant to the solution

= - \frac{1}{a ^{2} u} + C

\frac{d}{d x} (\frac{1}{sin ( x ^{2} )})

\int e^{4 x} cos (3 x) d x

\frac{\partial}{\partial x} (y + e^{x})

\int_{0}^{1} \frac{2 x}{1 + x ^{4}} d x

f (x) = \frac{- 4 x}{x ^{2} + 1}, at x = 0