Aus Online Mathematik Brückenkurs 2

Had the integral instead been

\displaystyle \int{e^{\sqrt{x}}}\centerdot \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 \frac{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}}\centerdot }}2\sqrt{x}\centerdot \frac{1}{2\sqrt{x}}\,dx \\ & =\left\{ \begin{matrix} u=\sqrt{x} \\ du=\left( \sqrt{x} \right)^{\prime }\,dx=\frac{1}{2\sqrt{x}}\,dx \\ \end{matrix} \right\} \\ & =\int{e^{u}\centerdot 2u\,du} \\ \end{align}

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

\displaystyle \begin{align} & \int{e^{u}\centerdot 2u\,du}=e^{u}\centerdot 2u-\int{e^{u}\centerdot 2\,du} \\ & =2ue^{u}-2\int{e^{u}\,du} \\ & =2ue^{u}-2e^{u}+C \\ & =2\left( u-1 \right)e^{u}+C \\ \end{align}

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

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

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.