$$\int \dfrac{x^2+x}{(e^x+x+1)^2}dx$$  My multiple attempts are as follows:-  Attempt $1$:  Let's differentiate $\dfrac{f(x)}{e^x+x+1}$  $$\dfrac{d}{dx}\left(\dfrac{f(x)}{e^x+x+1}\right)=\left(\dfrac{f'(x)(e^x+x+1)-f(x)(e^x+1)}{(e^x+x+1)^2}\right)$$   If we can choose $f(x)$ such that $f'(x)(e^x+x+1)-f(x)(e^x+1)$ equals to $x^2+x$, then our job will be done.  But I was not finding such $f(x)$.  Attempt $2$:  $$N_r=A(e^x+x+1)^2+2B(e^x+x+1)(e^x+1)$$ $$N_r=A(e^{2x}+x^2+1+2x+2xe^x+2e^x)+2B(e^{2x}+e^x+xe^x+x+e^x+1)$$ $$x^2+x=(A+2B)e^{2x}+(2A+4B)e^x+(2A+2B)xe^x+Ax^2+(2A+2B)x+A+2B$$  $$A+2B=0$$ $$A=1$$ $$2A+2B=0$$ $$2A+2B=1$$  So we don't have any solution for $A$ and $B$  Attempt $3$:  $$\int \dfrac{x^2+x+xe^x}{(e^x+x+1)^2}-\dfrac{xe^x}{(e^x+x+1)^2}dx$$ $$\int \left(\dfrac{(e^x+x+1)^2}{(e^x+x+1)^2}-\dfrac{1}{2}\cdot\dfrac{2(e^x+x+1)(e^x+1)}{(e^x+x+1)^2}\right)dx-\int \dfrac{xe^x}{(e^x+x+1)^2}dx$$  $$x-\ln(e^x+x+1)-\int \dfrac{xe^x}{(e^x+x+1)^2}dx$$  $$x-\ln(e^x+x+1)-\int \dfrac{(x+e^x+1-e^x-1)e^x}{(e^x+x+1)^2}dx$$  $$x-\ln(e^x+x+1)-\int \left(\dfrac{1}{e^x+x+1}-\dfrac{e^x+1}{(e^x+x+1)^2}\right)e^xdx$$  So we have the standard integral form $\int e^x\left(f(x)+f'(x)\right)=e^xf(x)+C$  $$x-\ln(e^x+x+1)-\dfrac{e^x}{e^x+x+1}+C$$  But it took me a lot of time to arrive at this. Is there any simple thing which I am missing here? Feel free to suggest your alternatives.
17272
Evaluate $$\int \frac{x^2+x}{(e^x+x+1)^2}dx$$    I tried converting in the form of Quotient rule(seeing the square in the denominator), neither am I able to make the denominators' derivative in the numerator. Some hints would be great. Thanks.
112982
Reasoning for $\frac{\ln|x|}{2}+ C$ :  $$\int  \frac{1}{(2x)}  \, dx = \frac{\ln|2x|}{2}+ C$$   Check: By taking derivative we get $\frac{1}{x}$  Reasoning for $\frac{\ln|x|}{2}+ C$ :  $$\int  \frac{1}{2x}  \, dx = \frac{1}{2}\int  \frac{1}{x}  \, dx =\frac{\ln|x|}{2}+ C$$  Check: By taking derivative we get $\frac{1}{x}$  Question: Is  $\int  \frac{1}{2x}  \, dx $ equal to $\frac{\ln|2x|}{2}+ C$ or $\frac{\ln|x|}{2}+ C$?
119685
Let $f(x)$  be a twice-differentiable function on $(a,b)$,show that   there exsit $\xi\in(a,b)$ ,such $$\int_{a}^{b}f(x)dx=\dfrac{1}{2}(b-a)[f(a)+f(b)]-\dfrac{1}{12}(b-a)^3f''(\xi)$$  if this problem condition is Amuss that $f(x)$ is a three-differentiable function,  $$f(x)=f(a)+f'(a)(x-a)+\frac{f''(\xi_{1})\cdot (x-a)^2}{2}$$ $$f(x)=f(b)+f'(b)(x-b)+\frac{f''(\xi_{2})\cdot (x-b)^2}{2}$$ so $(1)+(2)$ $$\Longrightarrow 2f(x)=f(a)+f(b)+f'(a)(x-a)+f'(b)(x-b)+\dfrac{1}{2}[f''(\xi_{1})+f''(\xi_{2})][(x-a)^2+(x-b)^2]$$ since $f(x)$ is a three-differentiable function,so $f''(x)$ is  continuous,so there exsit $\xi\in(\xi_{1},\xi_{2})$,such  $$\dfrac{1}{2}[f''(\xi_{1})+f''(\xi_{2})]=f''(\xi)$$ $$\int_{a}^{b}f(x)dx=\dfrac{1}{2}(b-a)[f(a)+f(b)]-\dfrac{1}{12}(b-a)^3f''(\xi)$$ But if $f(x)$ have twice-differentiable,this methods is not usefull  Then I use this methods can't prove it.
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 ^^
58934
Q)  $\int\sqrt{x+\sqrt{x^{2}+2}} dx$ .  Tried rationalising the numerator twice to get Numerator =-2 but not able to simplify denominator   The question reduces to (as per my rationalising)  $$\int \frac{-2}{\bigl(x - \sqrt{x^2+2}\bigr)\sqrt{x + \sqrt{x^2+2}}}\,dx$$
78427
I'm trying to find the value of the integral $\int_{-4}^{-2}e^{-x}\,dx$ but I just couldn't solve it.  Actually I found in a  that $\int e^x\,dx=e^x+C$ so I concluded:  $$ \int e^{-x}\,dx=\int\frac{1}{e^x}\,dx=\ln|e^x|$$  $$\int\limits_{-4}^{-2}e^{-x}\,dx=\left(\ln|e^{-2}|\right)-\left(\ln|e^{-4}|\right)=-2+4=2$$  I know the solution is wrong, but how can I solve this integral or any another integral like this?
101714
Let's say I have this function $f(x)=x$. I want to differentiate with respect to $x^2$. So I want to calculate $\large\frac{df(x)}{dx^2}$. In general, how can I calculate the derivative of a function $f(x)$ with respect to a function $g(x)$, so $\large\frac{df(x)}{dg(x)}$?   (I dont know whether this is a good notation)?
48420
How to prove $\int_{\mathbb{R}} e^{-\frac{(x+it)^2}{2}}dx=\sqrt{2\pi}$ for any $t\in \mathbb{R}$? I only obtained the case that $t=0$, $\int_{\mathbb{R}} e^{-\frac{x^2}{2}}dx=\sqrt{2\pi}$.   Thanks.
114177
We know that $$\int_0^1\frac{\ln (1+x)}{1+x^2}dx=\frac{\pi}8\ln 2,$$ but how about $$\int_0^1\frac{x\ln (1+x)}{1+x^2}dx?$$Prove that $$\int_0^1\frac{x\ln (1+x)}{1+x^2}dx=\frac{\pi^2}{96}+\frac{\ln^2 2}{8}.$$
19934
Integration is supposed to be the inverse of differentiation, but the integral of the derivative is not equal to the derivative of the integral:     $$\dfrac{\mathrm{d}}{\mathrm{d}x}\left(\int f(x)\mathrm{d}x\right) = f(x) \neq \int\left(\dfrac{\mathrm{d}}{\mathrm{d}x}f(x)\right)\mathrm{d}x$$   For instance: $$\begin{align*} &amp;\dfrac{\mathrm{d}}{\mathrm{d}x}\left(\int 2x+1\;\mathrm{d}x\right) &amp;&amp;= \dfrac{\mathrm{d}}{\mathrm{d}x}\left(x^2+x+C\right) &amp;= 2x+1\\ &amp;\int\left(\dfrac{\mathrm{d}}{\mathrm{d}x}\left(2x+1\right)\right)\mathrm{d}x &amp;&amp;= \int 2\;\mathrm{d}x &amp;= 2x+C\end{align*}$$  Why isn't it defined such that $\dfrac{\mathrm{d}}{\mathrm{d}x}a = \dfrac{\mathrm{d}a}{\mathrm{d}x}$, where $a$ is a constant, and $\int f(x)\;\mathrm{d}x = F(x)$? Then we would have: $$\begin{align*} &amp;\dfrac{\mathrm{d}}{\mathrm{d}x}\left(\int 2x+1\;\mathrm{d}x\right) &amp;&amp;= \dfrac{\mathrm{d}}{\mathrm{d}x}\left(x^2+x\right) &amp;= 2x+1\\ &amp;\int\left(\dfrac{\mathrm{d}}{\mathrm{d}x}\left(2x+1\right)\right)\mathrm{d}x &amp;&amp;= \int \left(2+\dfrac{\mathrm{d1}}{\mathrm{d}x}\right)\;\mathrm{d}x &amp;= 2x+1\end{align*}$$  Then we would have:     $$\dfrac{\mathrm{d}}{\mathrm{d}x}\left(\int f(x)\mathrm{d}x\right) = f(x) = \int\left(\dfrac{\mathrm{d}}{\mathrm{d}x}f(x)\right)\mathrm{d}x$$   So what is wrong with my thinking, and why isn't this the used definition?
36011
$$I_{n}=\int x^n(1-x^2)^{\frac{1}{2}} dx$$ Show that  $$(n+2)I_{n}=(n-1)I_{n-2}-x^{n-1}(1-x^2)^{\frac{3}{2}}$$  I can't seem to get this answer. Can someone please explain how to get to this?   Thanks in advance.
19430
What is the integral of $\frac{1}{x}$? Do you get $\ln(x)$ or $\ln|x|$? In general, does integrating $f'(x)/f(x)$ give $\ln(f(x))$ or $\ln|f(x)|$? Also, what is the derivative of $|f(x)|$? Is it $f'(x)$ or $|f'(x)|$?
102145
i'm required to evaluate this integral. I've tried factorizing but it doesn't lead me to anywhere.   $$\int\frac{x^2-1}{(x^4+3 x^2+1) \tan^{-1}\left(\frac{x^2+1}{x}\right)}\,dx$$  I've also tried letting $u = \frac{x^2+1}{x}$, $du/dx$ gets me $1-\frac{1}{x^2}$ but it doesn't seem to be working either.  Hope to receive some advise/ solutions on how to start tackling the question
9574
Integrate  $\displaystyle\int\dfrac{2x^2}{2x\cos(2x)+(x^2-1)\sin(2x)} \mathrm d x$   I tried dividing by $\cos^2(x)$ and then substituting $\tan(x)=t$.
119875
A friend asked me to prove     $$\int_0^1\frac{1-x}{(\ln x)(1+x)}\ dx=\ln\left(\frac2{\pi}\right)$$   and here is my approach:  \begin{align} I&amp;=\int_0^1\frac{1-x}{\ln x}\frac1{1+x}\ dx\\ &amp;=\int_0^1\left(-\int_0^1x^y\ dy\right)\frac1{1+x}\ dx\\ &amp;=\int_0^1\left(-\int_0^1\frac{x^y}{1+x}\ dx\right)\ dy\\ &amp;=\int_0^1\left((-1)^n\sum_{n=1}^\infty\int_0^1x^{y+n-1}\ dx\right)\ dy\\ &amp;=\int_0^1\left(\sum_{n=1}^\infty\frac{(-1)^n}{y+n}\right)\ dy\\ &amp;=\sum_{n=1}^\infty(-1)^n\int_0^1\frac1{y+n}\ dy\\ &amp;=\sum_{n=1}^\infty(-1)^n\left[\ln(n+1)-\ln(n)\right]\tag{1} \end{align}    Now how can we finish this alternating sum into $\ln\left(\frac2{\pi}\right)$?  My idea was to use   $$\operatorname{Li}_a(-1)=(1-2^{-a})\zeta(a)=\sum_{n=1}^\infty\frac{(-1)^n}{n^a}$$  and if we differentiate both sides with respect to $a$ we get  $$2^{1-a}(\zeta^{'}(a)-\ln2\zeta(a))-\zeta^{'}(a)=\sum_{n=1}^\infty\frac{(-1)^{n-1}\ln(n)}{n^a}\tag{2}$$  wolfram says that $\sum_{n=1}^\infty(-1)^n\ln(n)$ is divergent which means we can not take the limit for (2) where $a\mapsto 0$ which means we can not break the sum in (1) into two sums. So any idea how to do (1)?    Other question is, I tried to simplify the sum in (1) as follows  \begin{align} S&amp;=\sum_{n=1}^\infty(-1)^n\left[\ln(n+1)-\ln(n)\right]\\ &amp;=-\left[\ln(2)-\ln(1)\right]+\left[\ln(3)-\ln(2)\right]-\left[\ln(4)-\ln(3)\right]+\left[\ln(5)-\ln(4)\right]-...\\ &amp;=-2\ln(2)+2\ln(3)-2\ln(4)+...\\ &amp;=2\sum_{n=1}^\infty(-1)^n\ln(n+1) \end{align}  which is divergent again. what went wrong in my steps?  Thanks.
89004
That is:  $$\left(\int_{a(x)}^{b(x)}\!f(x,t)\,dt\right)&#39;$$  I don't know how to differentiate a integral if functions of $x$ are at its limits.  Can you guys show me how to do this?
13823
I'm trying to find $\int x^x \, dx$, but the only thing I know how to do is this:  Let $u=x^x$.  $$\begin{align} \int x^x \, dx&amp;=\int u \, du\\[6pt] &amp;=\frac{u^2}{2}\\[6pt] &amp;=\dfrac{\left(x^x\right)^2}{2}\\[6pt] &amp;=\frac{x^{2x}}{2} \end{align}$$  But it's certain that this isn't the correct way to evaluate that, and the answer must be wrong.