What is the integral of e^{r^2}r ?

The integral of e^{r^2}r is 1/2 e^{r^2}+C

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integral of e^{r^2}r

\int e^{r^{2}} r d r

\frac{1}{2} e^{r^{2}} + C

Solution steps

\int e^{r^{2}} r d r

Apply u-substitution

: \int \frac{e ^{u}}{2} d u

= \int \frac{e ^{u}}{2} d u

Take the constant out:

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

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

Use the common integral:

\int e^{u} d u = e^{u}

= \frac{1}{2} e^{u}

Substitute back

u = r^{2}

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

Add a constant to the solution

= \frac{1}{2} e^{r^{2}} + C

\frac{d}{d x} (ln (2 x^{2} + y^{2}))

25 y y^{^{'}} - 4 x = 0

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\int (x^{2} - 4)^{3} 2 x d x

\frac{- 12}{( 1 - x ) ^{4}}