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Integral

Yankel

Active member
Jan 27, 2012
398
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....

My incorrect answer would be

\[2\cdot \sqrt{x}\cdot e^{\sqrt{x}}\]


why isn't it ?
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,191
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
Jan 27, 2012
398
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
Jan 26, 2012
4,191
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 $x$.