I have seen something like this before: $\int \frac{dx}{(e+1)^2}$. This is apparently another way to write $\int \frac{1}{(e+1)^2}dx$.  However, considering this statement: $\int\frac{du}{(u-1)u^2} = \int du(\frac{1}{u-1}-\frac{1}{u}-\frac{1}{u^2})$. On the left side, $du$ is moved, If I had to evaluate an integral that is written in this way, how would I expand it into the usual $\int f(x)dx$ form?  (From the comments) Is this truly a product and if not why is it commutative?
27596
From reading answers here, I've noticed that some people write integrals as $\int dx \; f(x)$, while other people write them as $\int f(x)\;dx$.  I realize that there is no mathematical difference between the two notation forms, but was wondering why some people choose the first method over the second.  Is there some place in higher maths that it becomes beneficial to write the differential first?  (I, personally, have always used the second method, just because I was taught that way...)
8844
In an exercise, I'm supposed to assume that $|a| &lt; r &lt; |b|$ and prove that $\displaystyle\int_{\gamma} \frac{1}{(z - a)(z - b)} dz = \frac{2 \pi i}{a - b}$, where $\gamma$ is a circle of radius $r$ centered at the origin with positive orientation. So I had the idea to express this integral as $\frac{1}{a - b} \displaystyle\int_{\gamma} \frac{1}{z - a} - \frac{1}{z - b} dz$. Then I tried to evaluate each of these separately, but I don't really know what to do. I get $\displaystyle\int_{0}^{2 \pi} \frac{i r e^{it}}{r e^{it} - a} dt$ for the first term. How am I supposed to integrate this? I can't substitute $u = re^{it}$, can I?
100415
$$\int\frac{(\ln x)^{10}}{x}\,dx$$ All I know is that I am supposed to substitute $u=\ln x$. But can someone please explain to me how to find the anti derivative of $(\ln x)^{10}$. I think we are supposed to use integration by parts on integrating $\ln x$, but we haven't been taught that yet.
95678
could you help me on this integral ? $$\int \frac{1}{\cos(x)}\,\mathrm dx$$ Here's what I've started : $$\int \frac{1}{\cos(x)}\,\mathrm dx = \int \frac{\cos(x)}{\cos(x)^2}\,\mathrm dx = \int \frac{\cos(x)}{1-\sin(x)^2}\,\mathrm dx$$ Now, I did : $u = \sin(x)$, so $\mathrm du = 1$. Now I have : $$\int \frac{\text{???}}{1-u^2}\,\mathrm du$$ But at this point, I think I did the most of the job but I'm stuck. Could you help me to solve this integral please (to the integration by substitution at the end) ? Thanks EDIT : Now I follow the steps and I got : $$\int \frac{1}{1-u^2}\,\mathrm du$$ Doing the partial fraction I got $A = 1/2$ and $B = 1/2$. So basically I have : \begin{align} &amp; \int \frac{1}{1-u^2}\,\mathrm du = \int \frac{1/2}{1+u}\,\mathrm du + \int \frac{1/2}{1-u}\,\mathrm du \\[8pt] = {} &amp; \frac 1 2 \left(\int \frac{1}{1+u} \, du - \int \frac{1}{1-u} \, du\right) \\[8pt] = {} &amp; \frac 1 2 \ln\left(\frac{1+u}{1-u}\right) = \ln\left(\left(\frac{1+\sin(x)}{1-\sin(x)}\right)^{1/2}\right) \\[8pt] = {} &amp; \ln \left(\frac{\sqrt{1+\sin(x)}}{\sqrt{1-\sin(x)}}\right) = \ln\left(\frac{\sqrt{1+\sin(x)}}{\sqrt{\cos(x)^2}}\right) \\[8pt] = {} &amp; \ln\left(\frac{\sqrt{1+\sin(x)}}{\cos(x)}\right) \end{align} Here's what my teacher got :  What's wrong with what I did ? Did I miss something ?
86335
Question: How to evaluate $\displaystyle \int \frac{1}{\sin x\cos x} dx $  I know that the correct answer can be obtained by doing: $\displaystyle\frac{1}{\sin x\cos x} = \frac{\sin^2(x)}{\sin x\cos x}+\frac{\cos^2(x)}{\sin x\cos x} = \tan(x) + \cot(x)$ and integrating.  However, doing the following gets a completely different answer: \begin{eqnarray*} \int \frac{1}{\sin x\cos x} dx &amp;=&amp;\int \frac{\sin x}{\sin^2(x)\cos x} dx\\ &amp;=&amp;\int \frac{\sin x}{(1-\cos^2(x))\cos x} dx. \end{eqnarray*} let $u=\cos x, du=-\sin x dx$; then \begin{eqnarray*} \int \frac{1}{\sin x\cos x} dx &amp;=&amp;\int \frac{-1}{(1-u^2)u} du\\ &amp;=&amp;\int \frac{-1}{(1+u)(1-u)u}du\\  &amp;=&amp;\int \left(\frac{-1}{u} - \frac{1}{2(1-u)} + \frac{1}{2(1+u)}\right) du\\ &amp;=&amp;-\ln|\cos x|+\frac{1}{2}\ln|1-\cos x|+\frac{1}{2}\ln|1+\cos x|+C \end{eqnarray*}    I tested both results in Mathematica, and the first method gets the correct answer, but the second method doesn't. Is there any reason why this second method doesn't work?
59387
I some help computing   $$\int_{|z|=r} x \, dz$$  by noting that $x= \frac{1}{2}(z+\bar{z}) = \frac{1}{2}(z+\frac{r^{2}}{z})$ on the circle but I don't understand why this is true. Why does the complex conjugate of $z$ equal this on the circle?
48750
$$\begin{equation}     \int_{0}^{1}\int_{0}^{1}\int_{0}^{1} \left\{\frac{x}{y}\right\} \left\{\frac{y}{z}\right\}\left\{\frac{z}{x}\right\} dx\space dy\space dz \end{equation}$$  I am struggling with the above, I can deal with the case of double integrals but here due to my poor knowledge of triple integrals I am unable to decide what the limits should be in the transformed integral when I apply the substitution $\displaystyle \frac{x}{y}=u\space,\frac{y}{z}=v $ .  The integral turns to $\displaystyle \iiint \left\{u\right\}\left\{v\right\}\left\{\dfrac{1}{uv}\right\}z^2v\space du\space dv\space dz$ , but I have no idea about the limits.   Or even I am doing some mistake in these  steps only.   {.} Denotes Fractional Part
105433
Just as the product rule can be generalised to the product of more than two functions, i.e. $$\frac{d}{dx} \left [ \prod_{i=1}^k f_i(x) \right ]  = \sum_{i=1}^k \left(\frac{d}{dx} f_i(x) \prod_{j\ne i} f_j(x) \right) = \left(  \prod_{i=1}^k f_i(x) \right) \left( \sum_{i=1}^k \frac{f'_i(x)}{f_i(x)} \right),$$ is there a way to generalise integration by parts to evaluate $$\int f_1(x) f_2(x) \cdots f_n(x) dx \qquad ?$$    For contextual purposes only, I'm trying to evaluate by hand   $$\int x \cosh(x+1)e^x\sin(x) dx \quad  . $$  I know that I could let $u=x \cosh(x+1)$ and $v \ '=e^x\sin(x)$ but that would require integration by parts to be performed at least three times.  Is there a more-efficient way to evaluate this integral using the proposed 'generalisation'?  Thanks!
118451
My question is, how can I evaluate the following integral?  $$\int\frac1{\sqrt[4]{1+x^4}}\mathrm dx$$  Thanks.
80454
More specifically, I would like to know if there is a way I can express  $$\int \frac{x g'(x)}{f'(x)} dx $$  In terms of $f(x)$ and $g(x)$. Both $f(x)$ and $g(x)$ are non-negative and known to be increasing in $x$. If that helps.
28268
How to integrate $$\int \frac{v}{g-kv}\,\mathrm dv$$ it's supposed to be equal to $$ -\frac{v}{k} -\frac{g}{k^2}\ln(g-kv)$$ but I can't get that, I tried a substitution $u=g-kv$ and got close but no right? Any help.
70577
Compute the indefinite integral   $$ \int\frac{dx}{(x^2-x+1)\sqrt{x^2+x+1}} $$   My Attempt:  $$ \int\frac{dx}{(x^2-x+1)\sqrt{x^2+x+1}}=\int\frac{1}{(x^2-x+1)^{3/2}.\sqrt{\dfrac{x^2+x+1}{x^2-x+1}}}\,dx $$  Now define $t$ such that $t^2=\dfrac{x^2+x+1}{x^2-x+1}$ to get  $$ \begin{align} 2t\,dt &amp;= \frac{(x^2-x+1)(2x+1)-(x^2+x+1)\cdot (2x-1)}{(x^2-x+1)^2}\,dx\\ 2tdt &amp;= \frac{-4x^2+2x+2}{(x^2-x+1)^2}dx \end{align} $$  I don't know how to proceed from here.
85114
I'm having trouble computing the integral: $$\int \frac{\sin(x)}{\sin(x)+\cos(x)}\mathrm dx.$$ I hope that it can be expressed in terms of elementary functions. I've tried simple substitutions such as $u=\sin(x)$ and $u=\cos(x)$, but it was not very effective.  Any suggestions are welcome. Thanks.
117095
I tried finding the integral of $\int_{|z|=2}^{}\frac{1}{z^2+1}dz$ but not sure whether it is correct. $\gamma(t)=2e^{it},t\in[0,2\pi]$  $$\int_{|z|=2}^{}\frac{1}{z^2+1}dz=\int_{0}^{2\pi}\frac{2ie^{it}}{4e^{2it}+1}dt$$  I cannot seem to take it forward from here. Is it possible to use Cauchy's integral formula to get an answer? Thanks
90183
Evaluate the following integral.  $\int\frac1{(x+a)(x+b)}~dx$  ${a}\neq{b}$  I do not know where to being solving this integral.
69091
Ok, I give up, I have tried with $u$-substitution and integration by parts but I can't solve it. The integral is:  $$\int{\frac{e^x dx}{1+e^{2x}}}$$  I have tried $u=e^x$, $u=e^{2x}$ and also integration by parts but I can't solve it. The result should be:  $$\arctan(e^x)$$
33077
$\displaystyle \int \left( \frac{1}{x^2+3} \right)\; dx$  I've let $u=x^2+3$ but can't seem to get the right answer.  Really not sure what to do.