integrate from 0 to 1 $$\int \frac{x^7-1}{\log x} \, \mathrm{d}x$$![It's from MIT integration bee][1]  I try to solve this by substitution but nothing works out. Solving this problem from 10 days .I have no idea how they have a answer $\log(8)$.
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/A problem from the 2012 MIT Integration Bee is $$ \int_0^1\frac{x^7-1}{\log(x)}\mathrm dx $$ The answer is $\log(8)$. Wolfram Alpha gives an indefinite form in terms of the logarithmic integral function, but times out doing the computation. Is there a way to do it by hand?
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Evaluate $$I=\int_{0}^{1}\frac{x\log x}{\log(1-x)}dx$$   I try to change it to:$$I=\int_{0}^{1}\frac{(1-x)\log(1-x)}{\log x}dx$$ but I can't do next steps,help me ,thank you very much.
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We had our final exam yesterday and one of the questions was to find out the value of: $$\int_{0}^{\infty} \frac {\ln x}{1+x^2} \mathrm{d}x $$ Interestingly enough, using the substitution $x=\frac{1}{t}$ we get - $$-\int_{0}^{1} \frac {\ln x}{1+x^2} \mathrm{d}x = \int_{1}^{\infty} \frac {\ln x}{1+x^2} \mathrm{d}x $$and therefore $\int_{0}^{\infty} \frac {\ln x}{1+x^2} \mathrm{d}x = 0 $  I was curious to know about the theory behind this interesting (surprising even!) example.  Thank you.
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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?
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Answers to this question  assert that to find a solution to the differential equation $$\dfrac{dy}{dx} = \dfrac{y}{x}$$ we may rearrange and integrate $$\int \frac{1}{y}\ dy=\int \frac{1}{x}\ dx.$$  If we perform the integration we get $\log y=\log x+c$ or $$y=kx$$ for constants $c,k \in \mathbb{R}$.  I've seen others use methods like this before too, but I'm unsure why it works.  Question: Why is it legitimate to solve the differential equation in this way?
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$$\int_0^\infty x\left(\log\left(\frac1{x^2}\right)\right)^n dx$$  put $\log\left(\frac1{x^2}\right)=y^{1/n}$  $$I=\frac{\ln10}{2n}\int_{-\infty}^\infty10^{-y^{1/n}}y^{1/n}dy$$  put $10^{-y^{1/n}}=e^{-t}\implies$  $$I=\frac1{(2\ln10)^n}\int_{-\infty}^0e^{-t}t^ndt+\frac1{(2\ln10)}\int_0^\infty e^{-t}t^ndt$$  $$I=\frac1{(2\ln10)^n}\int_{-\infty}^0e^{-t}t^ndt+\frac{\Gamma(n+1)}{(2\ln10)}$$  I want to know how can I evaluate the first integral in the last step of my turn I used some substitutions so i get two integrals one of them converted to the formula of gamma function but the second could not be so because the domain of integration is different ? $n$ is a natural number
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$$\int^2_{-2}\frac{x^2}{1+5^x}$$  How do I start to integrate this?  I know the basics and tried substituting $5^x$ by $u$ where by changing the base of logarithm I get $\frac{\ln(u)}{\ln 5}=x$, but I got stuck.  Any hints would suffice preferably in the original question and not after my substitution.  (And also using the basic definite integrals property.)  Now I know only basic integration, that is restricted to high school, so would prefer answer in terms of that level.
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I've learned in my analysis class, that     $$ \int \frac{1}{x} \mathrm dx = \ln(x). $$   I can live with that, and it's what I use when solving equations like that. But how can I solve this, without knowing that beforehand.  Assuming the standard rule for integration is     $$ \int x^a \, \mathrm dx = \frac{1}{a+1} \cdot x^{a+1} + C .$$   If I use that and apply this to $\int \frac{1}{x} \,\mathrm dx$:  $$ \begin{align*}   \int \frac{1}{x}\mathrm  dx &amp;= \int x^{-1} \,\mathrm dx \\                       &amp;= \frac{1}{-1+1} \cdot x^{-1+1} \\                       &amp;= \frac{x^0}{0} \end{align*} $$  Obviously, this doesn't work, as I get a division by $0$. I don't really see, how I can end up with $\ln(x)$. There seems to be something very fundamental that I'm missing.  I study computer sciences, so, we usually omit things like in-depth math theory like that. We just learned that $\int \frac{1}{x} dx = \ln(x)$ and that's what we use.
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$$\int{\frac{1+x^2}{(1-x^2)\sqrt{1+x^4}}}\mathrm dx$$   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?
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I think I'm having a bad day. I was just trying to use integration by parts on the following integral:  $$\int \frac{dx}{x\log(x)}$$ Which yields $$\int \frac{dx}{x\log(x)} = 1 + \int \frac{dx}{x\log(x)}$$ Now, if I were to subtract  $$\int \frac{dx}{x\log(x)}$$ from both sides, it would seem that $0 = 1$. What is going on here?  Aside: I do know the integral evaluates to  $$\log(\log(x))$$ plus a constant.
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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.   $$\int\frac{s\sqrt{3}\sqrt{s^2-2ns\sqrt{7}+4n^2}}{2s^2-4ns\sqrt{7}+8n^2}dn$$
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I have been having extreme difficulties with this integral. I would appreciate any and all help. $$ \int \sqrt{\tan x} ~ \mathrm{d}{x}. $$
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I want to show that $\int_{0}^{\infty}\frac{\log x}{x^2+1}dx=0$, but is there a faster method than finding the contour and doing all computations?  Otherwise my idea is to do the substitution $x=e^t$, integral than changes to $\int _{-\infty}^{\infty}\frac{t e^t}{1+e^{2t}}dt$. Next step is to take the contour $-r,r,r+i\pi,-r+i\pi$ and integrate over it...
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$$\int \frac{\mathrm{d}x}{x}$$ If I integrate this by parts ($\displaystyle u=\frac{1}{x}, \mathrm{d}u = -\frac{\mathrm{d}x}{x^2}, \mathrm{d}v= \mathrm{d}x, v = x$), then why does this happen?  $$\int \frac{1}{x} \: \mathrm{d}x = u v - \int v \:\mathrm{d}u = \frac{x}{x} - \int x\left(\frac{-1}{x^2}\right) \:\mathrm{d}x=1+\int \frac{1}{x}\:\mathrm{d}x$$  So $\displaystyle\int \frac{1}{x}\:\mathrm{d}x = 1 + \int \frac{1}{x} \:\mathrm{d}x$ and cancel out both sides of $\displaystyle\int \frac{1}{x} \:\mathrm{d}x$ and I get $0=1$. How can I find the error in the process because right now I cant see it.
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How to prove the following   $$\int^1_0 \frac{\log\left(\frac{1+x}{1-x}\right)}{x\sqrt{1-x^2}}\,dx=\frac{\pi^2}{2}$$  I thought of separating the two integrals and use the beta or hypergeometric functions but I thought these are not best ideas to approach the problem.  Any other ideas ?
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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.
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Some time ago I came across  to the following integral:    $$I=\int_{0}^{1}\frac{1-x}{1+x}\frac{\mathrm dx}{\ln x}$$ What are the hints on how to compute this integral?