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Aus Online Mathematik Brückenkurs 2

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Version vom 10:28, 11. Mär. 2009

We shall solve the exercise in two different ways.

Method 1 (integration by parts)

At first sight, integration by parts seems impossible, but the trick is to see the integrand as the product

1lnx.

We integrate the factor 1 and differentiate lnx,

1lnxdx=xlnx−xx1dx=xlnx−1dx=xlnx−x+C.

Method 2 (substitution)

It seems difficult to find some suitable expression to substitute, so we try to substitute the whole expression u=lnx. The problem we encounter is how we should handle the change from dx to du. With this substitution, the relation between dx and du becomes

du=(lnx)dx=x1dx

and because u=lnx, then x=eu and we have that

du=1eudxdx=eudu.

Thus, the substitution becomes

lnxdx=udx=lnx=eudu=ueudu.

Now, we carry out an integration by parts,

ueudu=ueu−1eudu=ueu−eudu=ueu−eu+C=(u−1)eu+C

and the answer becomes

lnxdx=(lnx−1)elnx+C=(lnx−1)x+C.