+0  
 
0
319
5
avatar

integral of xe^square root of x dx

Guest Nov 29, 2014

Best Answer 

 #1
avatar+90968 
+13

 

After the substitution step I have used integration by parts 3 seperate times - How about that!

 

$$\\INTEGRATION BY PARTS\\\\
\boxed{\int uv'\;dx= uv-\int\;vu' \;dx}$$

 

$$\begin{array}{rllll}
\int x e^{\sqrt{x}}dx&=&\int\frac{2\sqrt{x}\;x\;e^{\sqrt{x}}}{2\sqrt{x}}\;dx\qquad\qquad &\left[\frac{d}{dx}x^{1/2}=\frac{1}{2}x^{-1/2}=\frac{1}{2\sqrt{x}}\right]\\\\
&=&2\int\frac{(\sqrt{x})^3e^{\sqrt{x}}}{2\sqrt{x}}\;dx\qquad\qquad &\\\\
&&&let\;t=\sqrt{x}\\\\
&&&\;dt=\frac{1}{2}x^{-1/2}dx=\frac{1}{2\sqrt{x}}dx\\\\
&=&2\int t^3e^tdt\qquad\qquad &\\\\
&=&2\left[t^3e^t-\int3t^2e^tdt\right] \qquad\qquad &u=t^3\quad v'=e^t\\
&& \qquad\qquad &u'=3t^2\quad v=e^t\\
&=&2t^3e^t-6\int t^2e^tdt \qquad\qquad &\\\\
&=&2t^3e^t-6\left[t^2e^t-\int 2te^tdt\right] \qquad\qquad &u=t^2\quad v'=e^t\\
&& \qquad\qquad &u'=2t\quad v=e^t\\
\end{array}$$

 

$$\begin{array}{rllll}
\int x e^{\sqrt{x}}dx &=&2t^3e^t-6t^2e^t+12\int te^tdt\right]\\\\
&=&2t^3e^t-6t^2e^t+12\left[te^t-\int e^t dt\right] \qquad\qquad &u=t\quad v'=e^t\\
&& \qquad\qquad &u'=1\quad v=e^t\\
&=&2t^3e^t-6t^2e^t+12te^t-12e^t+k\\\\
&=&2e^t\left[t^3-3t^2+6t-6\right]+k\\\\
&=&2e^{\sqrt{x}}\left[(\sqrt{x})^3-3(\sqrt{x})^2+6(\sqrt{x})-6\right]+k\\\\
&=&2e^{\sqrt{x}}\left[x^{3/2}-3x+6(\sqrt{x})-6\right]+k\\\\
\end{array}$$

Melody  Nov 29, 2014
Sort: 

5+0 Answers

 #1
avatar+90968 
+13
Best Answer

 

After the substitution step I have used integration by parts 3 seperate times - How about that!

 

$$\\INTEGRATION BY PARTS\\\\
\boxed{\int uv'\;dx= uv-\int\;vu' \;dx}$$

 

$$\begin{array}{rllll}
\int x e^{\sqrt{x}}dx&=&\int\frac{2\sqrt{x}\;x\;e^{\sqrt{x}}}{2\sqrt{x}}\;dx\qquad\qquad &\left[\frac{d}{dx}x^{1/2}=\frac{1}{2}x^{-1/2}=\frac{1}{2\sqrt{x}}\right]\\\\
&=&2\int\frac{(\sqrt{x})^3e^{\sqrt{x}}}{2\sqrt{x}}\;dx\qquad\qquad &\\\\
&&&let\;t=\sqrt{x}\\\\
&&&\;dt=\frac{1}{2}x^{-1/2}dx=\frac{1}{2\sqrt{x}}dx\\\\
&=&2\int t^3e^tdt\qquad\qquad &\\\\
&=&2\left[t^3e^t-\int3t^2e^tdt\right] \qquad\qquad &u=t^3\quad v'=e^t\\
&& \qquad\qquad &u'=3t^2\quad v=e^t\\
&=&2t^3e^t-6\int t^2e^tdt \qquad\qquad &\\\\
&=&2t^3e^t-6\left[t^2e^t-\int 2te^tdt\right] \qquad\qquad &u=t^2\quad v'=e^t\\
&& \qquad\qquad &u'=2t\quad v=e^t\\
\end{array}$$

 

$$\begin{array}{rllll}
\int x e^{\sqrt{x}}dx &=&2t^3e^t-6t^2e^t+12\int te^tdt\right]\\\\
&=&2t^3e^t-6t^2e^t+12\left[te^t-\int e^t dt\right] \qquad\qquad &u=t\quad v'=e^t\\
&& \qquad\qquad &u'=1\quad v=e^t\\
&=&2t^3e^t-6t^2e^t+12te^t-12e^t+k\\\\
&=&2e^t\left[t^3-3t^2+6t-6\right]+k\\\\
&=&2e^{\sqrt{x}}\left[(\sqrt{x})^3-3(\sqrt{x})^2+6(\sqrt{x})-6\right]+k\\\\
&=&2e^{\sqrt{x}}\left[x^{3/2}-3x+6(\sqrt{x})-6\right]+k\\\\
\end{array}$$

Melody  Nov 29, 2014
 #2
avatar+78557 
+5

Very clever, Melody........that initial "trick" of mutiplying the top and bottom by  2√x  was the key, wasn't it....???....I could have looked for hours and never discovered that......

(Maybe I should have bought some Coronas, too!!)

 

CPhill  Nov 29, 2014
 #3
avatar+90968 
+5

Yes maybe you should have    

Melody  Nov 29, 2014
 #4
avatar+78557 
0

Smart aleck.......

 

CPhill  Nov 29, 2014
 #5
avatar+26322 
+10

Here's an alternative method:

integration

.

Alan  Nov 29, 2014

10 Online Users

avatar
avatar
We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details