Integral of e Sqrt x

In this tutorial we shall explain another very important type of integral.

The integration of sine inverse is of the form
\[I = \int {{e^{\sqrt x }}dx} \,\,\,\,{\text{ – – – }}\left( {\text{i}} \right)\]

To solve this type of integration we use the method of integration by substitution.

If $$z = \sqrt x $$, taking its derivative gives
$$dz = \frac{1}{{2\sqrt x }}dx$$

and this gives
$$2zdz = dx$$

So integral (i) becomes
\[I = \int {2z{e^z}dz} \]

When using integration by parts it must have at least two functions. Here the first function is $$2z$$ and the second function is $${e^z}$$

Using the formula for integration by parts, we have
\[\int {\left[ {f\left( x \right)g\left( x \right)} \right]dx = f\left( x \right)\int {g\left( x \right)dx – \int {\left[ {\frac{d}{{dx}}f\left( x \right)\int {g\left( x \right)dx} } \right]dx} } } \]

Using the formula above, equation (i) becomes
\[\begin{gathered} I = 2z\int {{e^z}dz – \int {\left[ {\frac{d}{{dz}}2z\int {{e^z}dz} } \right]dz} } \\ \Rightarrow I = 2z{e^z} – 2\int {{e^z}dz} \\ \Rightarrow I = 2z{e^z} – 2{e^z} + c \\ \end{gathered} \]

Now again using the value $$z = \sqrt x $$, we have
\[\int {{e^{\sqrt x }}dx} = 2\sqrt x {e^{\sqrt x }} – 2{e^{\sqrt x }} + c\]