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