6
$\begingroup$

Integrals of the form $$\int{e^x(f(x)+f'(x))}dx$$ are very common. And I have seen this form appearing in several exam papers.But the problem I face with this particular type of integral is finding what $f(x)$ could be.Its more of a trial and error method.

My question is:

Is there any sure shot method to find $f(x)$(i.e. break the expression inside the bracket as $f(x)+f'(x)$) given that I know that the integral will be of the form $e^x(f(x)+f'(x))$?

$\endgroup$
11
$\begingroup$

Integrating by parts

$$\int e^x f'(x)dx=e^xf(x)-\int e^xf(x)dx.$$ Rearranging terms and adding a constant we get

$$\int e^x(f(x)+f'(x))dx=e^xf(x)+C.$$

Note that this equality holds for any differentiable function $f.$

Edit

If you want to write $g(x)$ as $f(x)+f'(x)$ you need to solve the differential equation $y'+y=g(x).$ Its solution is given by

$$y(x)=ce^{-x}+e^{-x}\int_1^x e^tg(t)dt.$$ That is, you need to know $\int e^xg(x)dx$ to solve it.

$\endgroup$
  • $\begingroup$ Sorry.But you did not understand my question I think.I know the formula you derived.But my question is that say if the integral is of the form $\int{e^x(g(x))}dx$ then how to split $g(x)$ as $\int{e^x(f(x)+f'(x))}dx$ ? $\endgroup$ – user220382 Feb 10 '16 at 16:44
  • $\begingroup$ I have edited the answer to clarify this point. $\endgroup$ – mfl Feb 10 '16 at 16:50
  • $\begingroup$ No it was not me who downvoted.I hardly ever downvote because everyone makes mistakes sometime or the other. $\endgroup$ – user220382 Feb 10 '16 at 16:53
  • $\begingroup$ OK, fair enough. At least I think you have the answer you look for now... $\endgroup$ – mickep Feb 10 '16 at 16:54
2
$\begingroup$

No, there is not a way such that for any given $g(x)$ which should be of he form $g(x) = f(x)+f'(x)$ to find the function $f(x)$ except trial and error and/or experience.

$\endgroup$
  • $\begingroup$ Not sure if he meant that. $\endgroup$ – Adrian Feb 10 '16 at 16:41
  • 1
    $\begingroup$ @adjan I meant that only.The others misinterpreted! $\endgroup$ – user220382 Feb 10 '16 at 16:44
  • $\begingroup$ How are you so sure @Hetebrij ? $\endgroup$ – user220382 Feb 10 '16 at 16:46
  • $\begingroup$ Well, in case you have $g(x) = x^5 + 5x^4$, it is easy. However, what about $g(x) = \log ( \log (x^2)^2) + \frac{ 4 }{\log(x^2)x} + \left( \frac{2}{x} \log(x) +1 \right) x^{ \log(x)}$? $\endgroup$ – Hetebrij Feb 10 '16 at 16:50
  • $\begingroup$ One can get $f$ by solving the ODE $y'+y=g(x).$ But this doesn't help for general $g.$ $\endgroup$ – mfl Feb 10 '16 at 16:52
1
$\begingroup$

You should know that $$ (\mathrm{e}^x)' = \mathrm{e}^x $$ So you can see that $$ \dfrac{d}{dx}\mathrm{e}^xf(x) = (\mathrm{e}^x)'f(x) + \mathrm{e}^xf'(x) $$ And then use the expression in my first line.

Beyond this I am not sure what the problem is?

$\endgroup$
  • 1
    $\begingroup$ I think you did not understand my question.... $\endgroup$ – user220382 Feb 10 '16 at 16:47
  • $\begingroup$ I know the formula you derived.But my question is that say if the integral is of the form $∫e^x(g(x))dx$ then how to split $g(x)$ as $∫e^x(f(x)+f′(x))dx$ ? $\endgroup$ – user220382 Feb 10 '16 at 16:47
1
$\begingroup$

I asked myself the same question as that by OP. I arrived at the fact that to find $f(x)$ one should solve the differential equation given above by mfl above. But the solution of the differential equation requires to compute first the integral of $g(x)e^x$, so a circular problem.

What OP is asking is how to solve the differential equation without doing integral calculus and when the solution $f(x)$ is found then we claim that the integral of $g(x)e^x$ is nothing but $f(x)e^x+C$.

To arrive at a satisfactory answer, do not ask yourself how to find a way to find $f(x)$ for ANY given $g(x)$. But ask how to find $f(x)$ for some classes of $g(x)$, as polynomials as so on. Classes we do find in textbooks exercises.

Actually, I did find a way for the following classes of $g(x)$ we find in textbooks.

  1. Polynomials (of any degree),

  2. sin x, cos x and any linear combinaison of sin x and cos x, next,

  3. $x \sin x$ and $x \cos x$ and any linear combinaison of them, next

  4. $x^{2}\sin x$ and $x^2\cos x$ and any linear combinaison of them, next

  5. Some partial fractions when the degree of the denominator is 2.

I also have tested the way to find $f(x)$ for each class cited above for most concrete examples in some textbooks (old and new). I do not believe that it is always better than usual integration by parts.

The same technique works for case $f'(x)+g'(x)f(x)$. Recall that $$\int (f'(x)+g'(x)f(x))e^{g(x)} dx = f(x)e^{g(x)}+C.$$

What i did discover of great value (at least for me) is that the $g(x)$ of ANY concrete textbook exercise is of the form $f(x)+f'(x)$ (or the form $f'(x)+g'(x)f(x)$) and the $f(x)$ can be found almost easily even for the following monster-looking example

Computing an awful integral

i did find the solution easily without trial-error method but by following a straight path!

When time permitted i will give some examples of finding $f(x)$ for $g(x)$ belonging to some of the cited classes.

EDIT: Here is a first example dealing with polynomial case

Evaluate the integral $\int \left( 2x^{5}+x^{3}+x\right) e^{x^{4}+x^{2}}dx$ It is of the form \begin{equation*} \int \left( 2x^{5}+x^{3}+x\right) e^{x^{4}+x^{2}}dx=\int h(x)e^{g(x)}dx. \end{equation*} This form recalls the well-known formula \begin{equation*} \int \left( f^{\prime }(x)+g^{\prime }(x)f(x)\right) e^{g(x)}dx=f(x)e^{g(x)}+C. \end{equation*} So we are done if we find a function $f(x)$ such that \begin{equation*} h(x)=f^{\prime }(x)+g^{\prime }(x)f(x). \end{equation*} In what follows, I will show that $f(x)=\frac{1}{2}x^{2},$ and therefore \begin{equation*} \int \left( 2x^{5}+x^{3}+x\right) e^{x^{4}+x^{2}}dx=\left( \frac{1}{2} x^{2}\right) e^{x^{4}+x^{2}}+C. \end{equation*} $\color{red}{\bf Problem:}$ We want to write $\left( 2x^{5}+x^{3}+x\right) $ as $ f^{\prime }(x)+g^{\prime }(x)f(x)$ where $g(x)=x^{4}+x^{2},$ $g^{\prime }(x)=4x^{3}+2x$ and $f(x)$ is to be determined. First, it is easy to see that \begin{equation*} \left( 2x^{5}+x^{3}+x\right) =x+(2x^{3}+x)(x^{2})=x+(4x^{3}+2x)(\frac{x^{2}}{ 2})=x+g^{\prime }(x)(\frac{x^{2}}{2}). \end{equation*} If we put $f(x)=(\frac{x^{2}}{2}),$ then \begin{equation*} f^{\prime }(x)+g^{\prime }(x)f(x)=\left( x\right) +(4x^{3}+2x)\left( \frac{ x^{2}}{2}\right) =x+2x^{5}+x^{3}=\left( 2x^{5}+x^{3}+x\right) , \end{equation*} we are done! Then, it suffices to take \begin{equation*} f(x)=(\frac{x^{2}}{2}).\ \ \ \color{red} \blacksquare \end{equation*}

EDIT: the following is of interest

$f(x) - f'(x) = x^3 + 3x^2 + 3x +1; f(9) =?$ EDIT: this one is of interest too Proof for formula $\int e^{g(x)}[f'(x) + g'(x)f(x)] dx = f(x) e^{g(x)}+C$

EDIT: I have provided an example here

Computing $\int (1 - \frac{3}{x^4})\exp(-\frac{x^{2}}{2}) dx$

EDIt another example here

Evaluating integral $\int\frac{e^{\cos x}(x\sin^3x+\cos x)}{\sin^2x}dx $

and The monster integral here

Compute $\int_0^{\pi/4}\dfrac{(1-x^2)\ln(1+x^2)+(1+x^2)-(1-x^2)\ln(1-x^2)}{(1-x^4)(1+x^2)} x\exp\dfrac{x^2-1}{x^2+1} dx$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy