What is the integral of (e^{-2}r)/a ?

The integral of (e^{-2}r)/a is (r^2)/(2e^2a)+C

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integral of (e^{-2}r)/a

integral

\frac{e ^{- 2} r}{a}

\frac{r ^{2}}{2 e ^{2} a} + C

Solution steps

\int \frac{e ^{- 2} r}{a} d r

Take the constant out:

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

= \frac{e ^{- 2}}{a} \cdot \int r d r

Simplify

\frac{e ^{- 2}}{a} : \frac{1}{e ^{2} a}

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

Apply the Power Rule

: \frac{r ^{2}}{2}

= \frac{1}{e ^{2} a} \cdot \frac{r ^{2}}{2}

Simplify

\frac{1}{e ^{2} a} \cdot \frac{r ^{2}}{2} : \frac{r ^{2}}{2 e ^{2} a}

= \frac{r ^{2}}{2 e ^{2} a}

Add a constant to the solution

= \frac{r ^{2}}{2 e ^{2} a} + C

1 - x

lo g_{10} (1 + x^{2})

x^{2} - x + 1

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

\frac{( 3 x ^{2} - 1 )}{( x ^{3} - x )}

What is the integral of (e^{-2}r)/a ?
The integral of (e^{-2}r)/a is (r^2)/(2e^2a)+C