What is the integral of 1/(a^2u^2) ?

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

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integral of 1/(a^2u^2)

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

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

Solution steps

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

Take the constant out:

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

= \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

f (x) = x^{2} - 4, (- 3, 5)

\frac{d y}{d t} = 1.5 y, y (0) = 100

\int_{0}^{\infty} t e^{- b t} d t

f (x) = - 8 x^{4} + 11 x^{2} - 3, (1, 0)

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