I need to compute this primitive :     $$ \int \frac{e^{3 x}}{x^3} dx$$   but I don't know how to proceed. I tried an integration by part and get     $$ \int \frac{e^{3x}}{x} dx$$   but then I am stuck... Is there no way else than making a McLaurin development ?  Thanks
89875
I made up some integrals to do for fun, and I had a real problem with this one. I've since found out that there's no solution in terms of elementary functions, but when I attempt to integrate it, I end up with infinite values. Could somebody point out where I go wrong?  So, I'm trying to determine: $$ \int{\frac{e^x}{x}} \, dx $$  Integrate by parts, where $u = 1/x$, and $v \, ' = e^x$. Then $u \, ' = - 1/x^2$, and $v=e^x$. So,  $$\int{\frac{e^x}{x}} \, dx = \frac{e^x}{x} + \int{\frac{e^x}{x^2}} \, dx$$  Integrate by parts again, $u = 1/x^2$, $v \, ' = e^x$, so that $u \, ' = -2/x^3$ and $v=e^x$. So,  $$\int{\frac{e^x}{x}} \, dx = \frac{e^x}{x} + \frac{e^x}{x^2} + 2\int{\frac{e^x}{x^3}} \, dx$$  Repeat this process ad infinitum to get,  $$\int{\frac{e^x}{x}} \, dx = \frac{e^x}{x} + \frac{e^x}{x^2} + 2 \left( \frac{e^x}{x^3} + 3 \left( \frac{e^x}{x^4} + 4 \left( \frac{e^x}{x^5} + \, \cdots \right) \right) \right) $$  Expanding this gives,  $$\int{\frac{e^x}{x}} \, dx = \frac{e^x}{x} + \frac{e^x}{x^2} + \frac{2e^x}{x^3} + \frac{6 e^x}{x^4} + \frac{24 e^x}{x^5} + \cdots $$  And factoring that gives,  $$\int{\frac{e^x}{x}} \, dx = \frac{e^x}{x} \left(  1 + \frac{1}{x} + \frac{2}{x^2} + \frac{6}{x^3} + \frac{24}{x^4} + \cdots \right) $$  Now, considering the series itself, the ratio between the $n^{th}$ term and the $(n-1)^{th}$ term = $\Large \frac{n}{x}$. Eventually, $n$ will be larger than $x$, so the ratio between successive terms will be positive, so (assuming $x$ is positive), the series diverges, meaning (and I'm sure everybody will cringe upon seeing notation used like this), that:  $$\int{\frac{e^x}{x}} \, dx = \frac{e^x}{x} \left( \infty \right) = \infty $$
64295
$$\int_{0}^{3} {(x^2+1)}d[x]$$ is equal to ___________ .  Attempt  $[x]$ is constant for every interval. So for that intervals $d[x]=0$ so intergral given is zero. Am I right?  Though the answer given is $\dfrac{17}{2}$.
19082
Compute the integral.   $$\int_{-\infty}^\infty \frac{x^4}{1+x^8} \, dx$$  The answer at the back of the book is  $$\frac{\pi}{4\sin(\frac{3\pi}{8})}$$
115290
\begin{eqnarray*} \int \frac{3\sin x+2\cos x}{2\sin x+3\cos x}dx &amp;=&amp;  \int \frac{(3\sin x+2\cos x)/\cos x}{(2\sin x+3\cos x)/\cos x}dx\\ \\ &amp;=&amp; \int \frac{3\tan x +2}{2\tan x +3} dx\\ &amp;&amp; u = \tan x \text{ and } du = \sec^2 x \ dx  \end{eqnarray*} Am I going in the right direction with this one? It seems like not.
11177
$ \int_{-2}^{0} \frac{x}{\sqrt{e^x+(x+2)^2}} dx $ = -$  2.887270... $  The function has no antiderivate and there's no symmetry here to help, if you solve this I would be thankful if you would for a highschooler ^^
116928
I was experimenting with various integration techniques, and I stumbled upon the integral:  $$ \int{\frac{\tan{x}}{x} dx} $$  I tried using a number of methods, but I could not solve it. Checking on WolframAlpha, I found that the integral does not have an elementary antiderivative. However, I am not satisfied with simply assuming it's non-elementary if I have difficulty in solving it.   My question: Is there a test to determine if an antiderivative is elementary or not?
113629
I came across this integral $\int e^{x^n} dx $ and was just wondering how you go about solving it. In this  there approach is to let $y = -x^n$, then $$ dy = -nx^{n-1} dx = -n(-y)^{\frac{n-1}{n}} \ dx \ \Rightarrow \ dx = -\frac{1}{n}(-y)^{\frac{1-n}{n}} \ dy, $$ then sub into the integral, we have $$ \int e^{x^n} dx = \frac{(-1)^{\frac{1}{n}-1}}{n}\int_{-\infty}^\infty  y^{\frac{1}{n}-1} e^{-y} \ dy, $$ then they referenced the primary definition of incomplete gamma function, $$ \Gamma(a,y) = \int_y^\infty t^{a-1} e^{-t} \ dt, $$ and claim that we can write our integral in terms of $\Gamma$ function, and the end result is $$ \int e^{x^n} dx = \frac{(-1)^{\frac{1}{n}-1}}{n} \Gamma(\frac{1}{n}, y). $$ Can anyone fill in the missing step for me? Our integral does look very similar to the gamma function, but what happened to the lower bound of the integral? In our integral the lower bound is $-\infty$; but to equate it to the gamma function we need the lower bound to be $y$.   Anyway substitute back the definition of $y$ we have the final result $$ \int e^{x^n} dx = \frac{(-1)^{\frac{1}{n}-1}}{n} \Gamma(\frac{1}{n}, -x^n). $$ I̶'̶m̶ ̶g̶u̶e̶s̶s̶i̶n̶g̶ ̶t̶h̶e̶i̶r̶ ̶a̶n̶s̶w̶e̶r̶ ̶i̶s̶ ̶c̶o̶r̶r̶e̶c̶t̶ ̶b̶e̶c̶a̶u̶s̶e̶ ̶M̶a̶t̶h̶e̶m̶a̶t̶i̶c̶a̶ ̶a̶l̶s̶o̶ ̶g̶i̶v̶e̶s̶ ̶t̶h̶e̶ ̶s̶a̶m̶e̶ ̶r̶e̶s̶u̶l̶t̶.̶ ̶B̶u̶t̶ ̶a̶n̶o̶t̶h̶e̶r̶ ̶t̶h̶i̶n̶g̶ ̶I̶ ̶d̶o̶n̶'̶t̶ ̶u̶n̶d̶e̶r̶s̶t̶a̶n̶d̶ ̶i̶s̶ ̶w̶h̶y̶ ̶d̶o̶e̶s̶ ̶t̶h̶e̶ ̶i̶n̶t̶e̶g̶r̶a̶l̶ ̶d̶e̶p̶e̶n̶d̶s̶ ̶o̶n̶ ̶$̶x̶$̶?̶ ̶Y̶o̶u̶'̶d̶ ̶t̶h̶i̶n̶k̶ ̶$̶x̶$̶ ̶s̶h̶o̶u̶l̶d̶ ̶b̶e̶ ̶i̶n̶t̶e̶g̶r̶a̶t̶e̶d̶ ̶o̶v̶e̶r̶.̶ Any help is appreciated, thanks.
92455
$$\int_{-\infty}^{\infty} \frac{\cos(αx)}{(x^2+1)(x^2+4)} \mathrm dx. $$  I am not sure how to solve this question. Can anyone help me to approach this problem. Thanks.
68108
I am having troubles with this integral:   $$\int \frac{1}{\sqrt{e^{2x}+e^x+1}}dx.$$   Could anyone help me?
88604
Essentially, all is in the title: Is it possible to compute the integral $$\int_1^{+ \infty} \frac{dx}{x^2(1+e^x)} \hspace{1cm} ?$$  I suspected some relation with a , but I was not able to find something explicit.
116570
This integral looks a lot like an elliptic integral, but with cubes instead of squares: $$I(a,b)=\int_0^\infty \frac{dx}{\sqrt{(x^3+a^3)(x^3+b^3)}}$$ Let's consider $a,b&gt;0$ for now. $$I(a,a)=\int_0^\infty \frac{dx}{x^3+a^3}=\frac{2 \pi}{3 \sqrt{3} a^2}$$ I obtained the general series solution the following way. Choose $a,b$ such that $a \geq b$, then: $$I(a,b)=\frac{1}{a^2} \int_0^\infty \frac{dt}{\sqrt{(t^3+1)(t^3+b^3/a^3)}}=\frac{1}{a^2} I \left(1, \frac{b}{a} \right)$$ $$\frac{b^3}{a^3}=p, \qquad I \left(1, \frac{b}{a} \right)=I_1(p)$$ $$I_1(p)=\int_0^\infty\frac{dt}{\sqrt{(t^3+1)(t^3+p)}}=2 \frac{d}{dp} J(p)$$ $$J(p)=\int_0^\infty\sqrt{\frac{t^3+p}{t^3+1}}dt=\int_0^\infty\sqrt{1+\frac{p-1}{t^3+1}}dt=$$ $$|p-1| \leq 1$$ $$=\sum_{k=0}^\infty \left( \begin{array}( 1/2 \\ ~k \end{array} \right) (p-1)^k \int_0^\infty \frac{dt}{(t^3+1)^k}$$ Now this is the most problematic part. The first integral of this series diverges. However, it's a constant in $p$, so if we differentiate, it formally disappears: $$I_1(p)=2 \sum_{k=1}^\infty \left( \begin{array}( 1/2 \\ ~k \end{array} \right) k (p-1)^{k-1} \int_0^\infty \frac{dt}{(t^3+1)^k}$$ Now, every integral in this series converges. The integtals can be computed using the Beta function, if we substitute: $$t^3=\frac{1}{u}-1$$ Finally, we rewrite: $$I_1(p)=\frac{\Gamma (1/3)}{3 \sqrt{\pi}} \sum_{k=1}^\infty \frac{k^2}{k!^2} \Gamma \left(k- \frac{1}{2}\right) \Gamma \left(k- \frac{1}{3}\right) (1-p)^{k-1}$$ Or, using the Pochhammer symbol:  $$I_1(p)=\frac{2 \pi}{3 \sqrt{3}} \sum_{k=0}^\infty \frac{(k+1)^2}{(k+1)!^2} \left(\frac{1}{2}\right)_k \left(\frac{2}{3}\right)_k (1-p)^k$$  My questions are:  Is the method I used valid (see the 'problematic part')? How to get this series into a Hypergeometric function form? Is there any 'arithmetic-geometric mean'-like transformation (Landen's transformation) for this integral? How to go about finding it?  If the method I used is correct, it can be used for any integral of the form ($m \geq 2$): $$I_m(a,b)=\int_0^\infty \frac{dx}{\sqrt{(x^m+a^m)(x^m+b^m)}}$$
91174
I was wondering how to determine $$\int x^2 e^{-x^2}\mathrm dx?$$ I tried to move the exponential into the integrator, and apply integration by parts, but that will make the integrand more complicated. Thank you!
50502
this is my first question here so I hope I did everything right. Still really new to LaTeX as well$$\int \frac{1}{x^3 -1}dx $$  I first used partial fractions to decompose this integral into two parts$$\int \frac{1}{3(x-1)}dx -\frac{1}{3} \int \frac{x+2}{x^2 + x + 1}dx$$   I definitely can solve the first part as $\frac{\ln(x-1)}{3}$ but I'm stuck with the second part as I don't know what to do since the denominator is not easily factor-able.   Thanks in advance to anybody taking the time to read this.  Edit: I do know that you can complete the square to make the integral prettier, but I don't know how to cancel the numerator or seperate it to relate the integral to $\int \frac{1}{x^2 + 1}$ to get some value of arctan(x) +c
65890
How do you integrate $\int \frac{1}{a + \cos x} dx$?  Is it solvable by elementary methods?  I was trying to do it while incorrectly solving a homework problem but I couldn't find the answer.  Thanks!
110587
I want to calculate the integral:  $$I \equiv \int_0^{\infty} \frac{dx}{1+x^3}$$  using residue calculus. I'm having trouble coming up with a suitable contour. I tried to take a contour in the shape of a quarter of a circle of radius $R$, then take the limit. The circular arc tends to zero, but the vertical fragment is problematic. I'm getting:  $$\int_0^{\infty} \frac{dx}{1+x^3} + \int_0^{\infty} \frac{idy}{1-iy^3} = w$$  where $w$ is the residue from the singularity at $e^{i \frac{\pi}{3}}$. If the second integral was purely imaginary, then it would be no problem, but it has a real part:  $$\int_0^{\infty} \frac{dx}{1+x^3} - \int_0^{\infty} \frac{x^3dx}{1+x^6} + i\int_0^{\infty} \frac{dx}{1+x^6} =w$$  So to get the answer I would have to know the value of the second integral, which doesn't look any easier than the first, and indeed I am unable to relate the two. Perhaps this approach is doomed, and there is a simpler way?
44933
What contour and integrand do we use to evaluate $$ \int_0^\infty \frac{x^3}{e^x-1} dx $$  Or is this going to need some other technique?
88127
Compute the indefinite integral  Irrational: $$\int \frac{x-1} {1 + \sqrt{x^2+2x-3}}dx$$ Help me pls. What do I need to do then? fraction isn't simplified