I just found something weird and I have no idea where it is wrong. I used two different approaches.  $\int _0^x(t^3-2t+3)\ dt=4\implies \frac{1}{4}x^4-x^2+3x=4\implies$ approximately $x=1.63$ or $x=-3.09$.  By the fundamental theorem of calculus, $\frac{d}{dx}\int _0^x(t^3-2t+3)\ dt=\frac{d}{dx}4\implies x^3-2x+3=0\implies$ approximately $x=-1.89$.   Could anyone tell me where it went wrong?
54709
Suppose we have $x^2+4x+3=0$  We differentiate both sides to get $2x+4=0$.  So $x=-2$  But $(-2)^2+4(-2)+3=0$   $4-8+3=-1$ is not equal to $0$.  Which tells that we cannot differentiate equalities. But in many proofs, like in integral calculus, differentiating both sides occurs. So, where is the fallacy?
95114
For two integration below, what is the ratio of them?  $$I_1 = \int_0^1\frac{dt}{\sqrt{1-t^4}}$$  $$I_2 = \int_0^1 \frac{dt}{\sqrt{1+t^4}}$$ What is the ration $\frac{I_1}{I_2}$? I do not have any thoughts for solving this question so could anyone give me some hints? Thank you!!
34537
What is the proof of the following: $$\int_{0}^{1} \left(\frac{\ln t}{1-t}\right)^2 \,\mathrm{d}t=\frac{\pi^2}{3} \&gt;?$$
22397
I want to compute this integral $$\displaystyle\int_{0}^{\infty} \frac{x^{1/3}}{1+x^{2}} \ dx$$  What I did was the following. I substituted $x=t^{6}$, so that my $dx= 6t^{5} \ dt$ and so the integral changes to $$\int_{0}^{\infty} \frac{t^2}{1+t^{12}} \cdot 6 t^{5} \ dt =6 \cdot \int_{0}^{\infty} \frac{t^7}{1+t^{12}} \: dt$$  Now If I substitute $t^{4}=v$ then what I will be having is the following integral $$\frac{6}{4}\cdot \int_{0}^{\infty} \frac{v}{1+v^{3}} \ dv$$  Now I can write $1+v^{3} = (1+v) \cdot (1-v+v^{2})$ and so I have   \begin{align*} \int_{0}^{\infty} \frac{v}{1+v^{3}}\: dv &amp;= \int_{0}^{\infty}\biggl[\frac{1}{3}\cdot \frac{v+1}{1-v+v^{2}} - \frac{1}{3} \cdot \frac{1}{1+v}\biggr]\: dv \end{align*}  Now the point is that the integral of $1/(1+v) \to \infty$, so I am not sure if this is the right way to do. Can anyone suggest anything?
19389
I have to compute this integral $$\int_{0}^{\infty} \frac{dt}{1+t^4}$$ to solve a problem in a homework. I have tried in many ways, but I'm stuck. A search in the web reveals me that it can be do it by methods of complex analysis. But I have not taken this course yet. Thanks for any help.
115119
Does anyone know the exact value of this integral?   $$\int^1_0 \frac{p^2-1}{\ln p}dp$$   I know that the approximate value is $1.09861...$. But I can't seem to get a figure on the exact value, or an antiderivative.
105184
Hi consider the following expression:  $F(x)=\int_a^{x}f(t)(x-t)dt$ and hence determine $F'(x)$.  This is what I have I done: first split into  $F(x)=\int_a^{x}xf(t)dt$ - $\int_a^{x}tf(t)dt$  The second half of the expression is integral of an expression fully interms of t, hence I know by the fundemental theorem of calculus upon differentiating it i can rewrite it as $xf(x)$.  However I have problem getting past the first half $\int_a^{x}xf(t)dt$ it is an integral containing x and t. is there any technique that I can apply in order to evaluate this integral?
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:  $$\int{\frac{e^x}{x}} \, dx = \frac{e^x}{x} \left( \infty \right) = \infty $$
102527
If  $\displaystyle f(x) = x^3-\frac{3}{2}x^2+x+\frac{1}{4}.$ then the value of $\displaystyle \int^{\frac{3}{4}}_{\frac{1}{4}}f(f(x))\mathrm dx$  $\displaystyle \int^{\frac{3}{4}}_{\frac{1}{4}}(f(x))^3-1.5(f(x))^2+f(x)+0.25 \mathrm dx$  could some help me with this, thanks
66971
I am told that an expression like $$ \int_a^x f(x)dx $$ is not well formed, i.e. it should be $$ \int_a^xf(t)dt $$ or similar.  Why is it that the limits of integration can't depend on the variable of integration?
90195
How to compute      $$\int_0^{+\infty} \frac{dt}{1+t^4} = \frac{\pi}{2\sqrt 2}.$$   I'm interested in more ways of computing this integral.   There is always the straight forward method to decompose into simple elements  $\dfrac{1}{(1 + t^{4})}$ it works but it is tedious. If someone has a faster and clever method, I'm interested :)  Update :   My quesion is different than  because all the solution that state there is talk about the the straight forward method to decompose into simple elements $\dfrac{1}{(1 + t^{4})}$ as i said before i'm not intersted in that way
1542
Calculating with Mathematica, one can have $$\int_0^{\pi/2}\frac{\sin^3 t}{\sin^3 t+\cos^3 t}\,\mathrm dt=\frac{\pi}{4}.$$   How can I get this formula by hand? Is there any simpler idea than using $u = \sin t$?     Is there a simple way to calculate $$ \int_0^{\pi/2}\frac{\sin^n t}{\sin^n t+\cos^n t}\,\mathrm dt $$ for $n&gt;3$? Could anyone come up with a reference for this exercise?
117243
Find out the value of $a, b$ and $c$, for which $$ \lim _{x\to 0}\:\:\frac{1}{x^5}\left(\int _0^x\:\left(a+b \cos\left(t\right)+c \cos\left(2t\right)\right)dt\right)=\frac{1}{5} $$  My first thought was to separate the integral into multiple ones because i thought that it is pretty clear that each integral is a common one and a,b,c are constants and can be taken out . After I solved the integral, that's where I had the problem. I couldn't do the limit of the result that I've got in order to find out the value of $a,b$ and $c$ .
94328
Using first principles, differentiate $$f(x)=(x)^{\frac{3}{4}} $$  So$$\lim_{h\to 0}\frac{f(x+h) -f(x)}{h} = \lim_{h\to 0}\frac{(x+h)^{\frac{3}{4}} -(x)^{\frac{3}{4}}}{h} $$  I cant seem to solve this question. I'm supposed to get the   output $\frac{3}{4}x^{-\frac{1}{4}} $
84841
There was a friend asking me how to prove  $$\int_0^\infty x^{-x}\,dx&lt;2$$  Mathematica showed that its approximate value is 1.99546, so I think it isn't easy to solve it, can you provide me some ideas about this question?
42702
Of course, it is easy to see, that the integral (or the antiderivative) of $f(x) = 1/x$ is $\log(|x|)$ and of course for $\alpha\neq - 1$ the antiderivative of $f(x) = x^\alpha$ is $x^{\alpha+1}/(\alpha+1)$.  I was wondering if there is an intuitive (probably geometric) explanation why the case $\alpha=-1$ is so different and why the logarithm appears?  Some answers which I thought of but which are not convincing:   Taking the limit $\alpha=-1$ either from above or below lead to diverging functions. Some speciality of the case $\alpha=-1$ are that both asymptotes are non-integrable. However, the antidrivative is a local thing, and hence, shouldn't care about the behavior at infinity.
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 $\dfrac{17}{2}$.