# Question 1f484

Dec 19, 2017

$\int \sqrt{x} {e}^{- \sqrt{x}} \mathrm{dx} = - 2 {e}^{- \sqrt{x}} \left(x + 2 \sqrt{x} + 2\right) + c$

#### Explanation:

For $\int \sqrt{x} {e}^{- \sqrt{x}} \mathrm{dx}$, let $u = \sqrt{x}$ and $\mathrm{du} = \frac{1}{2} \sqrt{x} \mathrm{dx}$

Then $\int \sqrt{x} {e}^{- \sqrt{x}} \mathrm{dx} = 2 \int {u}^{2} {e}^{u} \mathrm{du}$

For the integrand ${u}^{2} {e}^{- u}$, integrate by parts, $\int f \mathrm{dg} = f g - \int g \mathrm{df}$, where
f = u^2, dg = e^(-u) du, df = 2 u du, g = -e^(-u): 2intu^2e^udu = -2 e^(-u) u^2 + 4 int ue^(-u) du

For the integrand $u {e}^{- u}$, integrate by parts, $\int f \mathrm{dg} = f g - \int g \mathrm{df}$, where
f = u, dg = e^(-u) du, df = du, g = -e^(-u): -2 e^(-u) u^2 + 4 int ue^(-u) du = -4 e^(-u) u - 2 e^(-u) u^2 + 4 int e^(-u) du

The integral of ${e}^{- u}$ is -${e}^{- u} :$

$- 2 {e}^{- u} {u}^{2} + 4 \int u {e}^{- u} \mathrm{du} = - 2 {e}^{- u} {u}^{2} - 4 {e}^{- u} u - 4 {e}^{- u} + c$

Substitute back for u = sqrt(x): = -2 e^(-sqrt(x)) x - 4 e^(-sqrt(x)) sqrt(x) - 4 e^(-sqrt(x)) + c#

Which is equal to:

Answer: $= - 2 {e}^{- \sqrt{x}} \left(x + 2 \sqrt{x} + 2\right) + c$

Dec 19, 2017

$\int \sqrt{x} \left({e}^{-} \sqrt{x}\right) \mathrm{dx} = - 2 {e}^{-} \sqrt{x} \left(x + 2 \sqrt{x} + 2\right) + C$

#### Explanation:

We'll start this integration by using substitution.
For $\int \sqrt{x} \left({e}^{-} \sqrt{x}\right) \mathrm{dx}$, let $u = - \sqrt{x} = - {x}^{\frac{1}{2}}$.
Therefore, $\frac{\mathrm{du}}{\mathrm{dx}} \left[- {x}^{\frac{1}{2}}\right] = - \frac{d}{\mathrm{dx}} \left[{x}^{\frac{1}{2}}\right] = - \left(\frac{1}{2} {x}^{- \frac{1}{2}}\right) = - \frac{1}{2 \sqrt{x}}$, so $\mathrm{du} = - \frac{1}{2 \sqrt{x}} \mathrm{dx}$.
To substitute we need something that's part of the integrand, so rearrange this to $- 2 \sqrt{x} \mathrm{du} = \mathrm{dx}$

Now our integral looks like $\int \sqrt{x} \left({e}^{u}\right) \left(- 2 \sqrt{x} \mathrm{du}\right) = \int - 2 x {e}^{u} \mathrm{du}$.
Remember that $u = - \sqrt{x}$, therefore ${u}^{2} = x$ so our integral is equivalent to $\int - 2 {u}^{2} {e}^{u} \mathrm{du}$, which we can solve.

First apply the constant rule, noting that $\int - 2 {u}^{2} {e}^{u} \mathrm{du} = - 2 \textcolor{g r e e n}{\int {u}^{2} {e}^{u} \mathrm{du}}$

Now we need to use integration by parts, which states that $\int f \mathrm{dg} = f g - \int g \mathrm{df}$. Let $f = {u}^{2}$ and $\mathrm{dg} = {e}^{u} \mathrm{du}$. $\frac{\mathrm{df}}{\mathrm{du}} = \frac{d}{\mathrm{du}} \left[{u}^{2}\right] = 2 u$, so $\mathrm{df} = 2 u \mathrm{du}$. $\int \mathrm{dg} = \int {e}^{u} \mathrm{du} = {e}^{u}$, so $g = {e}^{u}$

$\textcolor{g r e e n}{\int {u}^{2} {e}^{u} \mathrm{du}} = {u}^{2} {e}^{u} - \int {e}^{u} \left(2 u \mathrm{du}\right) = {u}^{2} {e}^{u} - 2 \textcolor{p u r p \le}{\int u {e}^{u} \mathrm{du}}$

We need to use integration by parts again. This time, let $f = u$ and $\mathrm{dg} = {e}^{u} \mathrm{du}$. $\frac{\mathrm{df}}{\mathrm{du}} = \frac{d}{\mathrm{du}} \left[u\right] = 1$, so $\mathrm{df} = \mathrm{du}$. $\int \mathrm{dg} = \int {e}^{u} \mathrm{du} = {e}^{u}$, so $g = {e}^{u}$

$\textcolor{p u r p \le}{\int u {e}^{u} \mathrm{du}} = u {e}^{u} - \int {e}^{u} \mathrm{du} = u {e}^{u} - {e}^{u}$

$\textcolor{g r e e n}{\int {u}^{2} {e}^{u} \mathrm{du}} = {u}^{2} {e}^{u} - 2 \textcolor{p u r p \le}{\left(u {e}^{u} - {e}^{u}\right)} = {u}^{2} {e}^{u} - 2 u {e}^{u} + 2 {e}^{u}$

$- 2 \left(\textcolor{g r e e n}{{u}^{2} {e}^{u} - 2 u {e}^{u} + {e}^{u}}\right) = - 2 {u}^{2} {e}^{u} + 4 u {e}^{u} - 4 {e}^{u}$

Finally, plug $- \sqrt{x}$ back in for $u$ and simplify.

$- 2 x {e}^{-} \sqrt{x} - 4 \sqrt{x} {e}^{-} \sqrt{x} - 4 {e}^{-} \sqrt{x} = - 2 {e}^{-} \sqrt{x} \left(x + 2 \sqrt{x} + 2\right)$

We've omitted the constant of integration thus far since all we were doing was integration by parts and substitution, so we'll put it in now.

$- 2 {e}^{-} \sqrt{x} \left(x + 2 \sqrt{x} + 2\right) + C$