What is the integral of (r^2+r+1)e^r ?

The integral of (r^2+r+1)e^r is e^rr^2-e^rr+2e^r+C

Popular >

integral of (r^2+r+1)e^r

\int (r^{2} + r + 1) e^{r} d r

e^{r} r^{2} - e^{r} r + 2 e^{r} + C

Solution steps

\int (r^{2} + r + 1) e^{r} d r

Apply Integration By Parts

: (r^{2} + r + 1) e^{r} - \int (2 r + 1) e^{r} d r

= (r^{2} + r + 1) e^{r} - \int (2 r + 1) e^{r} d r

\int (2 r + 1) e^{r} d r = 2 e^{r} r - e^{r}

= (r^{2} + r + 1) e^{r} - (2 e^{r} r - e^{r})

Simplify

(r^{2} + r + 1) e^{r} - (2 e^{r} r - e^{r}) : e^{r} r^{2} - e^{r} r + 2 e^{r}

= e^{r} r^{2} - e^{r} r + 2 e^{r}

Add a constant to the solution

= e^{r} r^{2} - e^{r} r + 2 e^{r} + C

x \to 1 + lim (\frac{x ^{2} - 8 x + 15}{x - 1})

2 x y d x + (x^{2} + 1) d y = 3 x d x

\frac{\partial}{\partial r} (e^{r} cos (θ))

\int \frac{x ^{2} - 2 x + 1}{x ^{2} - 2 x + 10} d x

\int (2 x^{2} + 5) (x^{2} - 3 x) d x

What is the integral of (r^2+r+1)e^r ?
The integral of (r^2+r+1)e^r is e^rr^2-e^rr+2e^r+C