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}