What is the integral of (10)/(r^2) ?

The integral of (10)/(r^2) is -(10)/r+C

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integral of (10)/(r^2)

\int \frac{10}{r ^{2}} d r

- \frac{10}{r} + C

Solution steps

\int \frac{10}{r ^{2}} d r

Take the constant out:

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

= 10 \cdot \int \frac{1}{r ^{2}} d r

Apply the Power Rule

: - \frac{1}{r}

= 10 (- \frac{1}{r})

Simplify

10 (- \frac{1}{r}) : - \frac{10}{r}

= - \frac{10}{r}

Add a constant to the solution

= - \frac{10}{r} + C

6 t \frac{d y}{d t} + y = t^{3}

x^{2} + 4 y^{2}

y^{^{'}} = \frac{( 3 t ^{2} - e ^{t} )}{2 y - 5}

x \to 5 lim (2 e^{x} - 2 ln (x))

\frac{d}{d t} (- 2 t)