Question

Нужна помощь в решении неопределенных интегралов

Answer 1

$\displaystyle \int\dfrac{\sin(\arctan x)}{1+x^2}dx = \int \sin(\arctan x)d(\arctan x) = -\cos(\arctan x)+C$

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$\displaystyle \int\dfrac{x-2}{x^2-2x+2} = \dfrac{1}{2}\int\dfrac{d([x-1]^2)}{(x-1)^2+1}-\int \dfrac{d(x-1)}{(x-1)^2+1} = \dfrac{1}{2}\ln|(x-1)^2+1|-\arctan(x-1) =\dfrac{1}{2}\ln((x-1)^2+1)-\arctan(x-1)$

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$\displaystyle \int \dfrac{\sqrt[4]{x} }{1+\sqrt{x}}dx \stackrel{x=t^4}{=}\int\dfrac{t}{1+t^2}4t^3dt = 4\int\dfrac{t^4}{1+t^2}dt = 4\int (\tan(\arctan t))^4d(\arctan t) \stackrel{u=\arctan t}{=} 4\int \left(\dfrac{1}{\cos^2 u}-1\right)(\tan^2 u)du = 4\int\tan^2u d(\tan u) - 4\int \dfrac{1}{\cos^2 u}-1du = \dfrac{4}{3}\tan^3u-4\tan u +4u = \dfrac{4}{3}t^3 - 4t+4\arctan t +C = \dfrac{4}{3}x^{\frac{3}{4}}-4x^{\frac{1}{4}}+4\arctan x^{\frac{1}{4}}+C$