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integral of xe^square root of x dx

0
491
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integral of xe^square root of x dx

Guest Nov 29, 2014

#1
+92221
+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:

#1
+92221
+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
#2
+85817
+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
+92221
+5

Yes maybe you should have

Melody  Nov 29, 2014
#4
+85817
0

Smart aleck.......

CPhill  Nov 29, 2014
#5
+26640
+10

Here's an alternative method:

.

Alan  Nov 29, 2014

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