Lösung 2.3:2d
Aus Online Mathematik Brückenkurs 2
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
| \displaystyle 1\centerdot \ln x\,\textrm{.} |
We integrate the factor \displaystyle 1 and differentiate \displaystyle \ln x,
| \displaystyle \begin{align}
\int 1\cdot\ln x\,dx &= x\cdot\ln x - \int x\cdot\frac{1}{x}\,dx\\[5pt] &= x\cdot\ln x - \int 1\,dx\\[5pt] &= x\cdot\ln x - x + C\,\textrm{.} \end{align} |
Method 2 (substitution)
It seems difficult to find some suitable expression to substitute, so we try to substitute the whole expression \displaystyle u=\ln x\,. The problem we encounter is how we should handle the change from \displaystyle dx to \displaystyle du. With this substitution, the relation between \displaystyle dx and \displaystyle du becomes
| \displaystyle du = (\ln x)'\,dx = \frac{1}{x}\,dx |
and because \displaystyle u = \ln x, then \displaystyle x=e^u and we have that
| \displaystyle du = \frac{1}{e^u}\,dx\quad\Leftrightarrow\quad dx = e^u\,du\,\textrm{.} |
Thus, the substitution becomes
| \displaystyle \begin{align}
\int \ln x\,dx = \left\{\begin{align} u &= \ln x\\[5pt] dx &= e^u\,du \end{align}\right\} = \int ue^u\,du\,\textrm{.} \end{align} |
Now, we carry out an integration by parts,
| \displaystyle \begin{align}
\int u\cdot e^u\,du &= u\cdot e^u - \int 1\cdot e^u\,du\\[5pt] &= ue^u - \int e^u\,du\\[5pt] &= ue^u - e^u + C\\[5pt] &= (u-1)e^u + C\,, \end{align} |
and the answer becomes
| \displaystyle \begin{align}
\int \ln x\,dx &= (\ln x-1)e^{\ln x} + C\\[5pt] &= (\ln x-1)x + C\,\textrm{.} \end{align} |
