Question

Помогите взять интеграл от √(x^2+9)
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Answer 1

$int sqrt{x^2+9}, dx=[; x=3tgt; ,; dx= frac{3, dt}{cos^2t} ; ,; t=arctgfrac{x}{3}; ,=\\ =x^2+9=9tg^2t+9=9(tg^2+1)=9cdot frac{1}{cos^2t} , ]=int frac{frac{3}{cost}cdot 3, dt}{cos^2t} =\\=9, int frac{dt}{cos^3t} =9, int frac{sin^2t+cos^2t}{cos^3t}, dt=9, int frac{sin^2t, dt}{cos^3t}+9, int frac{dt}{cost}=A; ;\\ int frac{sintcdot sint, dt}{cos^3t}=[, u=sint; ,; du=cost, dt; ,; dv= frac{sint, dt}{cos^3t} ; ,\\v=-int frac{-sint, dt}{cos^3t}=-int frac{d(cost)}{cos^3t}=- frac{(cost)^{-2}}{-2} = frac{1}{2cos^2t} ; ]=$

$= frac{sint}{2cos^2t}-int frac{cost, dt}{2cos^2t} =frac{sint}{2cos^2t}- frac{1}{2} int frac{dt}{cost} ; ;\\A=frac{9, sint}{2cos^2t}-frac{9}{2}int frac{dt}{cost} +9int frac{dt}{cost} = frac{9, sint}{2cos^2t}+frac{9}{2}int frac{dt}{cost} ; ;\\star ; ; int frac{dt}{cost} =int frac{cost, dt}{cos^2t}=int frac{cost, dt}{1-sin^2t} =int frac{d(sint)}{1-sin^2t}=frac{1}{2}, lnBig |frac{1+sint}{1-sint} Big |+C; star \\A= frac{9sint}{2cos^2t}+frac{9}{4}, lnBig | frac{1+sint}{1-sint} Big |+C; ,; t=arctgfrac{x}{3}; .$