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Re: MX2 2015 Integration Marathon ALTERNATE METHOD: $ \begin{align*} \int_{0}^{\infty} \frac{2x^3-3x^2+1}{x^6-2x^3+x^2-2x+2} \text{ d}x &= \int_{0}^{\infty} \frac{(2x+1)(x-1)^2}{(x^3-1)^2+(x-1)^2} \text{ d}x \\ &= \int_{0}^{\infty} \frac{(2x+1)(x-1)^2}{(x-1)^2((x^2+x+1)^2+1)} \text{ d}x \\...

Re: MX2 2015 Integration Marathon NEW QUESTION: $ Evaluate $\int_{0}^{\infty} \frac{2x^3-3x^2+1}{x^6-2x^3+x^2-2x+2} \text{ d}x$ $

2013 Algebra Quiz 1 (Ver 2A & 2B) tests are now up! I will be posting solutions for them soon in the coming week. Link: https://www.dropbox.com/sh/q22hegzq1uu5uxt/AAB4rlHJkaWMsmaA6ML46HULa?dl=0 Enjoy! :)

Re: MX2 2015 Integration Marathon My bad, answer should then be \sin^{-1}x-\sqrt{1-x^2}+C :)

Re: MX2 2015 Integration Marathon Multiply the integrand by \sqrt{\frac{1+x}{1+x}} , then split the fraction into two components. This will get it into standard form, answer is \frac{\pi+2}{2} .

Re: MX2 2015 Integration Marathon Good job :). Answer could also be expressed as \frac{\pi}{2}\ln{|t|} .

Re: MX2 2015 Integration Marathon NEW QUESTION: $ Evaluate $\int_{\frac{1}{t}}^{t} \frac{\tan^{-1}x}{x}dx$ $

Re: UNSW Chit Chat Thread 2015. Officially out on 17 April. Solid timetable :)

Re: UNSW Chit Chat Thread 2015. Lol, I am guessing you thought ACCT1511 had like 37 hrs per week :p

Re: MX2 2015 Integration Marathon Damn that's neat! :)

Re: MX2 2015 Integration Marathon Don't know if this can be done within MX2 bounds..

Re: MX2 2015 Integration Marathon Almost there, I think you integrated the last part incorrectly, where the ln(u) comes from? You may have also mixed up some of the signs in your expression as well..

Re: MX2 2015 Integration Marathon NEW QUESTION: $ Evaluate $\int_{0}^{\infty} \frac{t^2}{(1+t)^5}dt$ $

Re: MX2 2015 Integration Marathon BUMP :)

Re: MX2 2015 Integration Marathon $ NOTE that $\pi^{x} = e^{(\ln{\pi})x}$, then use standard laws of integration of exponentials to get \\ $\frac{\pi^{x}}{\ln\pi}+C$ $

Re: MX2 2015 Integration Marathon :p

Re: MX2 2015 Integration Marathon NEW QUESTION: $ Evaluate $\int \frac{\sec\theta+\tan\theta}{\csc\theta+\cot\theta}d\theta$ $

Re: MX2 2015 Integration Marathon $ Use the substitution $u = \sqrt{x}$, then apply IBP twice to get \\ $2e^{\sqrt{x}}(x-2\sqrt{x}+2)+C$ $

Re: MX2 2015 Integration Marathon NEW QUESTION $ Find the exact value of $\int_{0}^{1} \frac{\ln(1+x)}{1+x^2}$ $

Re: MX2 2015 Integration Marathon $ Let $u^2 = 1+\sin(x)$, then \\ $\int \sqrt{1+\sin(x)}\cot(x)dx$ \\ = $\int \frac{2u^2}{u^2-1}du$ \\ = $2u + \ln\frac{u-1}{u+1}+C $ \\ = 2\sqrt{1+\sin(x)} + \ln\left(\frac{\sqrt{1+\sin(x)}-1}{\sqrt{1+\sin(x)}+1}\right)+C$ $