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x^{2}-x-6=0 -x+3\gt 2x+1 ; line\:(1,\:2),\:(3,\:1) f(x)=x^3 ; prove\:\tan^2(x)-\sin^2(x)=\tan^2(x)\sin^2(x) \frac{d}{dx}(\frac{3x+9}{2-x}) (\sin^2(\theta))' \sin(120)
24 kwi 2019 · I was wondering how you integrate $\arctan^2(x)$. I tried doing it by parts and allowing $u=\arctan(x)$ and $\frac{\mathrm dv}{\mathrm dx}=\arctan(x)$ . But from then it becomes complicated, I was wondering if there was alternative method, or different way of approaching the question.
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integral (arctanx)^2. Natural Language. Math Input. Extended KeyboardExamplesUploadRandom. Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals. For math, science, nutrition, history, geography, engineering, mathematics, linguistics, sports, finance, music…
$\begingroup$ According to Wolfram Alpha (integrals.wolfram.com/… ), you don't have an elementary indefinite integral. The answer involves the dilogarithm function.
Solution steps. ∫arctan(x2)dx. Apply Integration By Parts:xarctan(x2)−∫x4+12x2dx. =xarctan(x2)−∫x4+12x2dx. ∫x4+12x2dx=2(−421(ln2x2+22x+2−2arctan(2x+1))+421(ln2x2−22x+2+2arctan(2x−1))) =xarctan(x2)−2(−421(ln2x2+22x+2−2arctan(2x+1))+421(ln2x2−22x+2+2arctan(2x−1))) Add a constant to the ...
