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1) ∫ 1/2 arcsen 2y
P.F. Integrale
1/2 ∫ arcos 2y • d2x
∫ 1/2 arcos2t dt
Pon y= arcos2t • tg2y
∫ 1/2 tg2y dy = 1/2 tg2y - ∫ 1/2 x 2y y dy
Quindi
∫ 1/2 tg2y dy = 1/2 tg2y + 1/2 x2y - ∫ 1/2 tg2y dy
z/Σ x 1/2 tg2y = 1/2 tg2y + 1/2 (os2y + se2y) / 2
Quindi
∫ x arcosen2y dx = EX arcosen2y 2 (os (arcosen2x) + sen( arcosen2x)) / 2
+ C
1) ∫ ∗ x dx = ∫ x/1+x2 a xdx + ∫ x2 axdx
= 1t2 + c = ½ x2 axdx - ½ ∫ 1/x2 dx + c
= ∫ ax/1+x2 dx - ½ ∫ x2 axdx
= 1(1+x2)- ∫ ½/1+x2 dx + c.
∫ axdx = x2/1+ax - 1/2 + a xdx + c
∫ axdx = ²/1+x2 + a x/1 + ½ /1+x2 + ∫ a.
dx = ∫ dx/uln(x) = - ½ axdx + c
2) ∫ x2 dx = ∫ ax/1-x dln(x1) = a x2/2 + c
Quindi
∫ x dx + ½ x/1+x dx = 1x2/12+2 - ∫ x1&sub2; ax dx.
? x2/2 + 1 x + ax dx + a x2/1-c + c
Prova d'esame
Prof. Xovello
1. limx→∞ √5x2+2x4 - √x4+4
X e di precate una forma indeterminata ∞ - ∞
Quindi:
limx→∞ (•√5x2+2x4) - √x4+4 = limx→∞ (√x4+2x) - √x4
≤
limx→∞ [√x5 (x+2x) - √x4] = limx→∞ x [(x+2x)1/x - x (4+1/x)1/x]
So che:
limx→∞ [(1+x)1/x - 1 + ω]
ιω
1. 1/5 • 1/4 = 1/20
D: conjugate