Aus Online Mathematik Brückenkurs 2

Had the integral instead been

\displaystyle \int e^{\sqrt{x}}\cdot\frac{1}{2\sqrt{x}}\,dx

it is quite obvious that we would substitute \displaystyle u=\sqrt{x}, but we are missing a factor \displaystyle 1/2\sqrt{x} which would take account of the derivative of \displaystyle u which is needed when \displaystyle dx is replaced by \displaystyle du. In spite of this, we can try the substitution \displaystyle u=\sqrt{x} if we multiply top and bottom by what is missing,

\displaystyle \begin{align} \int e^{\sqrt{x}}\,dx &= \int e^{\sqrt{x}}\cdot 2\sqrt{x}\cdot \frac{1}{2\sqrt{x}}\,dx\\[5pt] &= \left\{\begin{align} u &= \sqrt{x}\\[5pt] du &= \bigl(\sqrt{x}\,\bigr)'\,dx = \dfrac{1}{2\sqrt{x}}\,dx \end{align}\right\}\\[5pt] &= \int e^{u}\cdot 2u\,du\,\textrm{.} \end{align}

Now, we obtain instead another, not entirely simple, integral, but we can calculate the new integral by partial integration (\displaystyle 2u is the factor that we differentiate and \displaystyle e^{u} is the factor that we integrate),

\displaystyle \begin{align} \int e^u\cdot 2u\,du &= e^u\cdot 2u - \int e^u\cdot 2\,du\\[5pt] &= 2ue^u - 2\int e^u\,du\\[5pt] &= 2ue^u - 2e^u + C\\[5pt] &= 2(u-1)e^u + C\,\textrm{.} \end{align}

If we substitute back \displaystyle u=\sqrt{x}, we obtain the answer

\displaystyle \int e^{\sqrt{x}}\,dx = 2(\sqrt{x}-1)e^{\sqrt{x}} + C\,\textrm{.}

As can be seen, it is possible to mix different integration techniques and often we need to experiment with different approaches before we find the right one.