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

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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

\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

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