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Sum
If u and v are two functions of x then prove that
`intuvdx=uintvdx-int[du/dxintvdx]dx`
Hence evaluate, `int xe^xdx`
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Solution
Let ` int vdx=w.....(1)`
`then " " (dw)/dx=v.....(2)`
`Now d/dx(u,w)=u.d/dx(w)+wd/dx(u)`
`=u.v+w(du)/dx......."from"(2)`
By definition of integration.
`u.w=int[u.v+w(du)/dx]dx`
`=intu.vdx+intw.(du)/dx dx`
`int u.v dx=u.w-int w (du)/dx dx`
`=u int v dx-int [(du)/dxintv.dx]dx`
[next section only required for question 2]
Hence, `int xe^xdx = x.inte^xdx-int[d/dx x.inte^xdx]dx`
`=xe^x-int1xxe^xdx`
`=xe^x-e^x+c`
Concept: Methods of Integration: Integration by Parts
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