All Questions
15
questions
1
vote
1
answer
79
views
How do we know the sign for the ln?
I used the $f'(x)/f(x)$ standard form to integrate the following:
$$\int \frac{\cot(x)}{\ln(\sin x)} \, \mathrm{d}x = \ln|\ln\sin(x)| + C$$
However, the correct answer involved removing the modulus ...
- 33
6
votes
2
answers
239
views
Prove that: $\sqrt [3]{36}<\ln 28<\sqrt [3]{37}$
Prove that:
$$\sqrt [3]{36}<\ln 28<\sqrt [3]{37}$$
This inequality is the result of an integral representation/inequality.
I lost access to the article that mentioned this inequality. Now I ...
- 1,659
0
votes
2
answers
53
views
Logarithms and Reciprocals
Show that the integral of $(e^x + e^{-x})/ (e^x -e^{-x})$ equals $\ln(1-e^{2x})-x+c$. I reached the stage where $\ln(e^{-2x} -1 ) +c$, but I don't know how to get the $-x$ and how to reverse the signs ...
1
vote
1
answer
95
views
Series expansion for logarithm
Find the Taylor series for
$$
f(x)=\int^x_1 \ln\left(2t^2 - 4t+11\right)\,dt \text{, expanded at about } x_0=1
$$
and find the radius of convergence of the series.
My approach: First, I found the ...
- 406
1
vote
1
answer
114
views
Substitution that gives a division by $0$
When solving the following integral
$$ \int_{0}^{c}\frac{\ln(1+x^{2})}{x^{2}}dx $$
(where c is a real number; it's a constant that really doesn't matter for the problem anyway), after doing ...
- 75
0
votes
2
answers
44
views
In an integral, why does logarithmic function of an exponential completely drop out?
How to show that
$$-\int_{-\infty}^{\infty} e^{-(x-\mu)^2 / 2\sigma^2} \ln\left[ e^{-(x-\mu)^2 / 2\sigma^2} \right] \mathrm{d} x$$
equals this:
$$-\int_{-\infty}^{\infty} e^{-(x-\mu)^2 / 2\sigma^2} ...
- 1,554
3
votes
0
answers
112
views
Evaluate $\int_{0}^{1} \frac{\ln(1-x)\ln^2(1+x)\:dx}{x}$ [duplicate]
Evaluate $$I=\int_{0}^{1} \frac{\ln(1-x)\ln^2(1+x)\:dx}{x}$$
We have $$\frac{\ln(1-x)}{x}=-\sum_{k=1}^{\infty}\frac{x^{k-1}}{k}$$
Hence $$I=-\sum_{k=1}^{\infty}\left(\frac{1}{k}\int_{0}^{1}x^{k-1}\...
- 10.2k
0
votes
3
answers
48
views
Algebra problem in integration by parts
The integral to solve:
$$
\int{5^{sin(x)}cos(x)dx}
$$
I used long computations using integration by parts, but I don't could finalize:
$$
\int{5^{sin(x)}cos(x)dx} = cos(x)\frac{5^{sin(x)}}{ln(5)}+\...
user748992
-1
votes
3
answers
75
views
Solve the differential equation $x^2y_2=2y$
Solve the differential equation $$x^2y_2=2y$$
My try:
Let $$z=x^2y_1$$
Differentiating with respect to $x$ we get
$$z_1=x^2y_2+2xy_1=x^2y_2+\frac{2z}{x}$$
$\implies$
$$z_1-\frac{2z}{x}=2y$$
So ...
- 10.2k
2
votes
4
answers
359
views
Why does $\ln(2+\sqrt3)={1\over2}\ln(7+4\sqrt3)$?
I was solving the definite integral $\int_{\sqrt7}^{2\sqrt7}{1\over\sqrt {x^2-7}}dx$, and came out with the intermediate step $\int_{\sqrt7}^{2\sqrt7}\sec\theta\ d\theta$, which led me to finish off ...
- 945
1
vote
2
answers
169
views
Find the value of $\int_{0}^{1} \frac{x \log x \:dx}{1+x^2}$
Find the value of $$I=\int_{0}^{1} \frac{x \log x \:dx}{1+x^2}$$
My Try: I used Integration by parts
So
$$I=\frac{1}{2}\log x \times \log (1+x^2) \biggr\rvert_{0}^{1}-\int_{0}^{1}\frac{\log(1+x^2)}{...
- 13.2k
0
votes
3
answers
162
views
Find the indefinite integral $\int \frac{\ln( x)} {x(1-\ln (x))}\,dx $ [closed]
Evaluate$$\int \frac{\ln (x)}{x(1-\ln(x))}\,dx$$
2
votes
4
answers
2k
views
Integral $\int {t+ 1\over t^2 + t - 1}dt$
Find : $$\int {t+ 1\over t^2 + t - 1}dt$$
Let $-w, -w_2$ be the roots of $t^2 + t - 1$.
$${A \over t + w} + {B \over t+ w_2} = {t+ 1\over t^2 + t - 1}$$
I got $$A = {w - 1\over w - w_2} \qquad B = {...
user312097
0
votes
1
answer
83
views
Logarithm simplification from a double integral question
I solved this integral
$$\int_0^{ln2} \int_{e^y}^2 \frac xydxdy$$
and got this: $$\frac{-\ln(2)^2}8 -\frac{\ln(2)}8 + \frac{3}{16} $$
However, when I checked the answer from the back of the book, ...
- 111
6
votes
1
answer
317
views
Problems with ln(ax) equations.
After fiddling around with the ln() function, I arrived at a problem.
I have found that $a \approx 1.39095$. However, I couldn't find the exact value.
Using the Lambert w function, I have already ...
