# Integral

#### Yankel

##### Active member
Hello,

I have this integral here:

$\int e^{\sqrt{x}}dx$

and I wanted to ask, why can't I treat it like I would treat this integral:

$\int (3x+5)^{5}dx$

In which I would integrate as if g(x)=3x+5 is a normal x, and then divide by the inner derivative ? I tried it with the upper integral, and it doesn't work, the solution includes another "-1" which I don't understand where comes from....

$2\cdot \sqrt{x}\cdot e^{\sqrt{x}}$

why isn't it ?

#### Ackbach

##### Indicium Physicus
Staff member
The first integral is of an exponential function (variable exponent), and the second is of a power function (variable base). They are completely different functions, behave differently, and have quite different antiderivatives. To do $\int e^{ \sqrt{x}} \, dx$, I would probably try by-parts first, and see what you get.

#### Yankel

##### Active member
but

$\int e^{3x}dx$

does work using the method I specified. How do I know when I can use and when I can't ?

(I mean the method of integrating and then dividing by the inner derivative).

#### Ackbach

##### Indicium Physicus
Staff member
I suppose your method would work any time the inner function of the exponential or power function is linear (or, more correctly, affine). That is, for any integral of the following types:
\begin{align*}
&\int e^{mx+b} \, dx, \quad \text{or} \\
&\int (mx+b)^n \, dx.
\end{align*}
But $\sqrt{x}$ is not an affine function of $x$.