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Question
If u and v ore differentiable functions of x. then prove that:
`int uv dx = u intv dx - int [(du)/(d) intv dx]dx`
Hence evaluate `intlog x dx`
Solution
Let `intv dx`= w ......(i)
Then, `(dw)/(dx)` = v ......(ii)
Now, `d/(dx)(u, w) = u. d/(dx)(w) + w d/(dx)(u)`
= `u.v + w (du)/(dx)` ......[From (ii)]
By definition of integration
u.w = `int[u.v + w . (du)/(dx)]dx`
∴ u.w = `intu.v dx + intw. (du)/(dx) dx`
∴ `intu.v dx = u.w - intw. (du)/(dx) dx`
= `uintv dx - int[(du)/(dx) intv. dx]dx` ......[Using (i)]
Hence, `intlogx dx = xlogx - int 1/x x xx dx`
= x log x – x + C
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