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Calculation of \(\displaystyle{ \displaystyle \int^{2}_{1}\frac{\sqrt{x^4+1}}{x^2}dx}\)

Thanks Premislav. Can we solve it without gamma function?

<r><QUOTE author="Premislav" post_id="5588770" time="1570116086" user_id="99778"><s>[quote=Premislav post_id=5588770 time=1570116086 user_id=99778]</s> <LATEX><s>[latex]</s>\int\frac{(x^2+x+1)^3}{\sqrt{1-x^2}}dx=\left|\begin{array}{ccc}\sqrt{1-x^{2}}=xt+1\\x=-\frac{2t}{t^{2}+1}\\\mbox{d}x =\frac{2\l...

Thanks arek1357

\(\displaystyle{ \int\frac{(x^2+x+1)^3}{\sqrt{1-x^2}}dx}\)

Thanks Premislav.

Finding \(\displaystyle{ \int^{1}_{0}\frac{\arctan\bigg(\frac{x}{x+1}\bigg)}{\arctan\bigg(\frac{1+2x-x^2}{2}\bigg)}dx}\)

Thanks Arek.

\(\displaystyle{ \displaystyle \lim_{n\rightarrow \infty}\frac{\int^{\frac{\pi}{2}}_{0}(\sin x+\cos x)^{n+1}dx}{\int^{\frac{\pi}{2}}_{0}(\sin x+\cos x)^{n}dx}}\)

Thanks premislav. \displaystyle \int\frac{x^2(x\sec x+\tan x)}{(x\tan x-1)^2}dx -- 9 May 2019, o 10:57 -- If \displaystyle I_{k}=\int^{e}_{1}\frac{1}{(1+\ln x)^k}dx, Then \displaystyle \sum^{\infty}_{k=0}(1_kI_{k}) -- 9 May 2019, o 10:59 -- Thanks premislav. \displaystyle \int\frac{x^2(x\sec x+\tan ...

Thanks ptemislav

-- 27 Apr 2019, o 17:10 --

\(\displaystyle{ \displaystyle \int\frac{(x^2-1)\sin x+x\cos x}{(x^2+1)-2x\sin (2x)+(x^2-1)\cos(2x)}}\)

Thanks premislav

Evaluation of \(\displaystyle{ \displaystyle \int^{2\pi(n^2+1)}_{2\pi n^2}\frac{1}{\bigg[(a^2+b^2)+(a^2-b^2)\cos x)\bigg]^2}dx}\) where \(\displaystyle{ a,b>0,n\in \mathbb{N}}\)

Thanks Arek

Calculate If \(\displaystyle{ \displaystyle I = \int_{0}^{\pi}\frac{\sin (884 x)\sin (1122x)}{2\sin x}dx}\) and \(\displaystyle{ \displaystyle J = \int_{0}^{1}\frac{x^{238}(x^{1768}-1)}{x^2-1}dx\;,}\) Then \(\displaystyle{ \displaystyle \frac{I}{J}}\)

Thanks premislav and arek \displaystyle \int_0^\infty \frac{x}{e^x-1}\,dx=\int_0^\infty \frac{xe^{-x}}{1-e^{-x}}\,dx \displaystyle =\sum_{n=0}^\infty \int_0^\infty xe^{-(n+1)x}\,dx \displaystyle =\sum_{n=1}^\infty \frac1{n^2} \displaystyle =\frac{\pi^2}{6} Evaluate \displaystyle \int^{\frac{\pi}{4}}...

\(\displaystyle{ \displaystyle \int^{2\pi}_{0}x^2\sec(x)dx}\)-- 29 Mar 2019, o 09:42 --\(\displaystyle{ \displaystyle \int^{\infty}_{0}\frac{x\ln x}{1+e^x}dx}\) and \(\displaystyle{ \displaystyle \int^{\infty}_{0}\frac{x\sin x}{(x^2+1)^3}dx}\)

Thanks friends got it Let \displaystyle I(\alpha) = \int^{2}_{0}\frac{\ln(1+\alpha x)}{1+x^2}dx\;\;, \alpha>0 Then  \displaystyle I'(\alpha) = \int^{2}_{0}\frac{x}{(1+\alpha x)(1+x^2)}dx = \frac{1}{1+\alpha^2}\int^{2}_{0}\bigg(\frac{-\alpha}{1+\alpha x}+\frac{\alpha +x}{1+x^2}\bigg)dx So  \displayst...