Here's the integral:  $$\int_{1}^{5}{\frac{\sqrt{x}}{\sqrt{6-x}+\sqrt{x}}dx}$$  tried rationalising the function:  $$\frac{\sqrt{x\left( 6-x \right)}-x}{6-2x}$$  which doesn't get me anywhere with substitution or integration by parts    I would appreciate if someone were to help me through hints/techniques (besides numerical methods).
123753
Evaluate:  $$\int_{1}^{2011} \frac{\sqrt{x}}{\sqrt{2012 - x} + \sqrt{x}}dx$$  Using real methods only.  I am not sure what to do.  I tried finding a power series, which was too ugly.  I just need some hints, not an answer to do this integral, this is from the MIT Integration bee 2012.
117001
Evaluate $$\int \frac{2-3x}{2+3x}\sqrt\frac{1+x}{1-x}d x$$ What substitution should I use ? $\sqrt\frac{1+x}{1-x}$ suggests $x=cos2\theta $ but its not useful in $\frac{2-3x}{2+3x}$
102137
$$\int \frac{x^2}{(x\cos x -\sin x)(x\sin x +\cos x)}\ dx$$   My approach :   Dividing the denominator by $\cos^2x$ we get $\dfrac{x^2\sec^2x }{(x -\tan x)(x\tan x +1)}$ then  $$\int \frac{x^2\sec^2x}{x^2\tan x -x\tan^2x+x-\tan x}\ dx=\int \frac{x^2(1+\tan^2x)}{x^2\tan x -x\tan^2x+x-\tan x}dx$$  But I am not getting any relation between numerator and denominator so that I will take any substitution and solve further please suggest whether it is correct and how to proceed in this. Thanks.
95025
Find the indefinite integral  $$\int {dx \over {(1+x^2) \sqrt{1-x^2}}} $$  Is there a smart substitution or algebric trick that I'm missing? Because integration by parts hasn't helped..
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?
1209
Evaluate the integral $$\int_0^{\pi}\frac{a^n\sin^2x+b^n\cos^2x}{a^{2n}\sin^2x+b^{2n}\cos^2x}dx.$$   I have no idea.   $$\int_0^{\pi}\dfrac{a^n\sin^2x+b^n\cos^2x}{a^{2n}\sin^2x+b^{2n}\cos^2x}dx=\int_0^{\pi/2}\dfrac{a^n\sin^2x+b^n\cos^2x}{a^{2n}\sin^2x+b^{2n}\cos^2x}dx=\int_0^{\pi/2}\dfrac{a^n\tan^2x+b^n}{a^{2n}\tan^2x+b^{2n}}dx$$  I try the substitution $\tan{x}=t$, but it doesn’t work.
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:  $$\arctan(e^x)$$
4558
Show that $$\int\limits_0^\pi{\frac{{{x^2}}}{{\sqrt 5-2\cos x}}}\operatorname d\!x =\frac{{{\pi^3}}}{{15}}+2\pi \ln^2 \left({\frac{{1+\sqrt 5 }}{2}}\right).$$   I don't have any idea how to start, but .
51644
I know that $\int{\frac{1}{x}}dx$ is simply $\ln{(x)}+c$ (-which is clearly unrelated to the problem but I just thought I would share anyway) but I am not sure how to approach $e^{x{^2}}$. Perhaps a substitution?
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.
68815
I have to evaluate:  $$\int_{0}^{\pi/2}\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}\, \mathrm{d}x. $$ I can't get the right answer! So please help me out!
109343
I want to compute the following integral to show that  $$ \int_0^\infty \left(\frac{1}{x}\right)^{\ln(x)}\,\mathrm{d}x=\mathrm{e}^{\frac{1}{4}}\sqrt{\pi}.$$ I have no idea where the $\sqrt{\pi}$ comes from.     Since the indefinite integral can't be evaluated with elementary functions I'm stuck.   I am just curious.
30247
So, the question says I have to perform the indefinite integration $$\int\frac{e^x(2-x^2)}{(1-x)\sqrt{1-x^2}}dx$$ I know that $$\int e^x(f(x)+f'(x))dx=e^xf(x)+C$$ Since any other substitution (using $x=ln(t)$ etc.) doesn't work, I expect I have to break the fraction with $e^x$ in the above integral to make separate fractions for $f(x)$ and $f'(x)$ somehow. I separate $2-x^2=1+(1-x^2)$, but that doesn't work (leaves $x$ in the numerator of derivative which I don't see in the question). Any other tricks?
106957
$$\int_0^\pi x \ln(\sin (x))dx  $$   I tried integrating this by parts but I end up getting integral that doesn't converge, which is this $$ \int_0^\pi \dfrac{x^2\cos (x)}{\sin(x)} \ dx$$ So can anyone help me on this one?
66145
I have a calculus II final coming up and this question came up in a past final exam:  $$\int_{6}^{16} \frac{dx}{\sqrt{x^3 + 7x^2 + 8x - 16}} = \frac{\pi}{k},$$  where $k$ is a constant. Find $k$.  My progress so far:  $$\int_{6}^{16} \frac{dx}{\sqrt{(x - 1)}(x + 4)} = \frac{\pi}{k}$$  The answer is: $k = 6\sqrt{5}$  I do not know where to from from this step. Any helps or hints will be greatly appreciated! Thank you.  EDIT: $(x + 4)$ not $(x + 2)$ in the denominator.
32843
By integration by parts and the substitution $x = \sin t$ we can easily calculate the integral $\int_{0}^{1} \ln (x+ \sqrt{1-x^2})dx$ which equals to $\sqrt{2} \ln (\sqrt{2} +1) -1.$ I’ve tried to use the same substitution $x = \sin t$ to calculate the integral $ \int_{0}^{1} \frac {\ln (x+ \sqrt{1-x^2})}{x}dx,$ which becomes $$ \int_{0}^{\frac {\pi}{2}} \frac {\ln \sin (t+ \frac {\pi}{4})}{\sin t}dt$$ It seems difficult to solve the particular integral. Any help?
96422
I came across the following integral, and I don't know how to solve it. $$ \int\log\left(x+\sqrt{x^2-1}\right)dx $$ I tried the "obvious" substitution of $x=\sec\theta$, which gives you: $$ \int\tan\theta\sec\theta\log\left(\tan\theta+\sec\theta\right)d\theta $$ However, this doesn't simplify it much, in the sense that I have no idea how to solve this now! Take out the common factor of $\sec$ from the log, or perhaps do $u=\tan\theta+\sec\theta$, but both of these lead to dead ends (at least for me).  In case you are wondering, Wolfram|Alpha claims the answer is: $$x\log\left(x+\sqrt{x^2-1}\right)-\sqrt{x^2-1}+c$$