I don't understand how the integral   $$\int \frac{\sqrt{9-x^2}}{x}=\sqrt{9-x^2}-3 \ln(\sqrt{9-x^2}+3)+3 \log(x)+c$$  I keep getting $-3/x +c$ as the answer.
122862
I need to solve $\int \frac{\sqrt{25-y^2}}{y}dy$.  I originally thought IBP, but that led to a very large and confusing algebra problem.   Then I started to look at the $\sqrt{25-y^2}$ and started to consider doing a $u$-substituion, and letting $u=\sin\theta$ and $du=\cos\theta d\theta$. Once I started playing a bit, the intuition was lost.  Any and all help is appreciated.
33079
This integral has been bugging me since yesterday:     $$\int_0^1\frac{\ln\ln\left({1}/{x}\right)}{x^2-x+1}\mathrm dx$$   I've tried substitution $y={1}/{x}$ and $e^y={1}/{x}$, but those didn't help much. Wolfram Alpha gives me result: $-0.67172$. Could anyone here please help me to obtain the closed form of the integral preferably (if possible) with elementary ways (high school methods)? Any help would be greatly appreciated. Thank you.
389
The integral is from P. Nahin's "Inside Interesting Integrals...", problem C2.1. His proposed solution includes trigonometric substitution and the use of log-sine integral.  However, I think the problem should have an easier solution (without appealing to another complicated integral at least).  I have the following trick in mind. Let's introduce the substitution $x=4-z$  $$I=\int_0^4 \frac{\ln x}{\sqrt{4x-x^2}}~dx=\int_0^4 \frac{\ln (4-z)}{\sqrt{4z-z^2}}~d(4-z)=\int_0^4 \frac{\ln (4-z)}{\sqrt{4z-z^2}}~dz$$  $$2I=\int_0^4 \frac{\ln (4x-x^2)}{\sqrt{4x-x^2}}~dx$$  $$I=\int_0^4 \frac{\ln \sqrt{4x-x^2}}{\sqrt{4x-x^2}}~dx$$    And here I'm stuck. I'm not sure if this can go somewhere. Maybe partial integration can help, but I don't know how to choose the functions. What do you think?    Here is .  Only  does not use trig substitution, it used gamma function instead. If there are no other ways, I'm prepared to give up on my question. But I would be grateful if it's left open at least for several days    Edit  After many attempts, I conclude that there is no trick to this integral. The reason is: the general form of this integral in not zero, but has the same symmetry properties, as the above case:  $$I(a)=\int_0^a \frac{\ln x}{\sqrt{ax-x^2}}~dx=\int_0^a \frac{\ln (a-x)}{\sqrt{ax-x^2}}~dx=\int_0^a \frac{\ln \sqrt{ax-x^2}}{\sqrt{ax-x^2}}~dx \neq 0$$  $$I(4)=0$$  So we will get nothing from symmetry considerations alone. There are two possible ways to solve this - either trigonometric substitution or gamma function.  Edit 2  I was wrong it seems, see the accepted answer.
107906
I have to verify that     $$\int_0^\pi \ln(1+\alpha\cos(x))dx=\pi\ln\left(\frac{1+\sqrt{1-\alpha^2}}{2}\right)$$   with $|\alpha|&lt;1$. It is my homework and don't know where to begin.
68428
I have no idea how to find this integral:  $$\int_0^1\frac{(x-x^2)^4}{1+x^2}\ dx\ ?$$  Any help would be appreciated. Thanks in advance.
32720
$${\dfrac{1}{\pi^2}}\int_{0}^{\infty}\dfrac{{(\ln{x}})^2}{\sqrt{x}{(1-x)^2}} \mathrm d x$$   I tried substituting $1/x$ for $x$ but the the only change in the integral is that the $\sqrt{x}$ moves in the numerator from the denominator. I don't understand what to substitute. $\tan{x}$ doesn't seem to work.
85817
How do I solve this? $$\int_0^\infty\frac{\ln(1+{x^2})}{1+{x^2}}$$ I know the answer is $\pi\ln2$.
108140
How can i integrate following definite integral?  I tried integrating by parts which obviously did not work.  $$I=\int_0^1 \sqrt[3]{x\log{\frac{1}x}} \, \mathrm dx$$
62548
I am trying to evaluate $$I=\int_{1/2}^1 \frac{\log(x)}{\sqrt{x^2+1}}\,dx$$ My attempt was to integrate by parts with $u=\log(x)$ and $v=\log{(x+\sqrt{1+x^2}})$ to get$$I=-\log{\frac{1}{2}}\log{\left(\frac{1}{2}+\sqrt{1+\frac{1}{4}}\right)}-\int_{1/2}^1 \frac{\log{(x+\sqrt{1+x^2}})}{x}\,dx$$ so simplifying the integral is $$I=\log{(2\phi)}-I_1$$ where $I_1=\int_{\frac{1}{2}}^1 \frac{\sinh^{-1}(x)}{x}\,dx$ If possible I would love to get some help in evaluating those.  I also thought about considering $I(k)=\int_{\frac{1}{2}}^1 \frac{x^k} {\sqrt{x^2+1}}\,dx$ writting the denominator as a binomial series, integrate it then derivate and plug $k=0$ to get the answer but the series is not nice.
30836
Calculate   $$\frac{\int_0^\pi x^3\ln(\sin x)\,dx}{\int_0^\pi x^2\ln(\sqrt{2}(\sin x))\,dx}$$   In this problem, I'm unable to understand how to start.  I tried applying integration by parts but I couldn't solve it. I also tried the various properties of definite integration but they were of no use. Maybe applying integration by parts (or DI method) successively may work but it leads to a form of $\frac{\infty}{\infty}$.
89872
How do you go about integrating this? For context purposes, $s$ is a constant and $n$ is just the variable we need to integrate for. If needed, a number can be substituted for $s$, but I would like it without preferably.   $$\int\frac{s\sqrt{3}\sqrt{s^2-2ns\sqrt{7}+4n^2}}{2s^2-4ns\sqrt{7}+8n^2}dn$$
87553
How to figure out the following integral? I have not been able to solve it from some time. $$\int_0^1\frac{x^{2} -1}{\log x}\,dx$$
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)  $$\int \frac{-2}{\bigl(x - \sqrt{x^2+2}\bigr)\sqrt{x + \sqrt{x^2+2}}}\,dx$$
69861
$$\int{\frac{1+x^2}{(1-x^2)\sqrt{1+x^4}}}\mathrm dx$$   This was a Calc 2 problem for extra credit (we have done hyperbolic trig functions too, if that helps) and I didn't get it (don't think anyone did) -- how would you go about it?
12417
How can I find the value of this following definite integral?     $$\int_2^4\frac{\sqrt{\log(9-x)}}{\sqrt{\log(9-x)}+\sqrt{\log(3+x)}}dx$$
52766
How to solve the following integral $$\int_{0}^{\infty} \frac{ \ln x}{(x+a)(x+b)} dx,$$ where $a,b&gt;0$ and $a \neq b$. I was looking for some kind of substitution. However, I don't see an obvious one.  Thanks!
54135
The integral to be solved is $\int \sqrt{x^8 + 2 + x^{-8}} \,\mathrm{d}x$. I tried substitution $t=x^8$ which got me to $\frac{1}{8} \int \sqrt{t + 2 + \frac{1}{t}} t^{-7/8} \,\mathrm{d}t$ and I'm stuck. Can you help?