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integral of xe^square root of x dx

Guest Nov 29, 2014

Best Answer 

 #1
avatar+93317 
+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
 #1
avatar+93317 
+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+88871 
+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+93317 
+5

Yes maybe you should have    

Melody  Nov 29, 2014
 #4
avatar+88871 
0

Smart aleck.......

 

CPhill  Nov 29, 2014
 #5
avatar+26971 
+10

Here's an alternative method:

integration

.

Alan  Nov 29, 2014

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