What is the integral of 7re^{r/4} ?

The integral of 7re^{r/4} is 7(4e^{r/4}r-16e^{r/4})+C

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integral of 7re^{r/4}

\int 7 r e^{\frac{r}{4}} d r

7 (4 e^{\frac{r}{4}} r - 16 e^{\frac{r}{4}}) + C

Solution steps

\int 7 r e^{\frac{r}{4}} d r

Take the constant out:

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

= 7 \cdot \int r e^{\frac{r}{4}} d r

Apply Integration By Parts

: 4 e^{\frac{r}{4}} r - \int 4 e^{\frac{r}{4}} d r

= 7 (4 e^{\frac{r}{4}} r - \int 4 e^{\frac{r}{4}} d r)

\int 4 e^{\frac{r}{4}} d r = 16 e^{\frac{r}{4}}

= 7 (4 e^{\frac{r}{4}} r - 16 e^{\frac{r}{4}})

Add a constant to the solution

= 7 (4 e^{\frac{r}{4}} r - 16 e^{\frac{r}{4}}) + C

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

f (x) = x^{2} - 4, y = 0, x = - 4, x = 3

y = x^{2} - 6, (4, 10)

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

5 x^{2} + x y + 5 y^{2} = 11, (1, 1)

What is the integral of 7re^{r/4} ?
The integral of 7re^{r/4} is 7(4e^{r/4}r-16e^{r/4})+C