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Example 16 Find jinx dx, (x > 0). We cannot apply the logarithmic rule here, because that rule deals with the integrand 1 /x, not In x. Nor can we use Rule VI. But if we let v = In x, implying dv = (1 /x) dx, and also let u = x, so that du = dx, then the integration can be performed as follows: J In x dx = jv du = uv — ju dv

= xln x — jdx = x In x — x + c = x(ln x — 1) + c

Example 17 Find Jxex dx. In this case, we shall simply let v = x, and u = ex, so that dv = dx and du = ex dx. Applying Rule VII, we then have jxex dx = jv du = uv — Ju dv

The validity of this result, like those of the preceding examples, can of course be readily checked by differentiation.

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