Analisi 1, correzione esame

1)

z̅−1/2=z²+2

|x+iy−4|²=(x+iy)²+2

(x−4)²+y²=x²+2ixy−y²+2

x²−2x+4+y²=x²−y²+2

2y²=−2x+2

3 punti distinti

2)

1. ∫₁^∞ 4x² dx / (3x⁴+2x³−1)

2. ∫₀^∞ 4/3 dt / (3t³+2t−2)

−∫ 4/3 dt / (t+1)(3t−1)

= −3∫ [ln|t+1| − 1/4 ln|3t−1|] + c

= −3∫ [ln|x³+1| + 1/3 ln|3x²+1| + c]

1)

z - 1² = z² + 2

z = x +iy

|x + iy - 1|² = (x + a(y)²) + 2

(x - a)² + y² = x² + 2ixy - y² + 2

x² - 2x + y² = x² - y² + z

( x D(x)\)

\( \Rightarrow \) non puó convergere

. se \(x=0 \, \sum_{n=1}^{+\infty}\frac{n\sqrt{n}}{n^2+1}\) diverge \( \Rightarrow A_n\, \) diverge

\(A_n\sim \frac{n}{n} \)

. se \(x> 0 \sim \frac{n}{n^2exn} ne^x \) \(\rightarrow\) serie geometrica che converge

\(\sqrt{x} \, \text{esterno} \, \text{del} \, \text{comparto}\)

\( \Rightarrow 0 \, \text{para converge} \, x>0 \)

\(*\text{potrei} \, \text{usare} \, \, \text{outli}\)

\(\text{no} \, \text{caricata} \, \text{del} \,\text{radica}\)

5) \(\int_{+\infty}^0 \frac{3x}{4x^{n}+4x^2-3}dx \)

        \(2x^2=t \, \rightarrow \, dt=4xdx \)

          \(\int \frac{3x}{4x^{n}+4x^2-3}\,dx = \)

    \(=\frac{3}{4}\int \frac{dt}{(t+3)(t-3)} \)

     \(\frac{1}{(t+3)(t-r)}\,\, \frac{At+A_t+3B+3B}{(t+3)(t-r)(r-t)}\)

        \(t=x^2\)

4t+u__t=3

t-2+4; 1+2 \, \text{A}\)

\(-\frac{3}{2} \)

\(t=\frac{3}{2} \) \( \Rightarrow \frac{1}{2} \)

\( \frac{1}{(t+\frac{3}{2})(t-\frac{1}{2})}\)

\((2t+3)(2t+3)= \)

\((2x^2+3)(2x^2-1)\)

\( \left\{n+0=0\right\}

{\(-A+3B-4=2\)

\(\B=2/4\)

\{A=-2/4\}

= ³/₄ ∫_-1⁺¹ (-t/4 + t/4)/ (t+3)(t-1) dt

= 3(-ln|t+3| + ln|t-1|) + C = ^{3 ln |2x2-1|}/_{16 [2x+3]} + C

X → +∞

X = 3

X → 0

X = 3

= ^{3 ln 5}/₁₆

integrale gia risolto

6) ^+∞ ∑ _n=1 ^a_n/_{n³ + |x|}

per quali valori di x converge?

- se |x| > 1 allora a_n non tende ad 0 No convergenza

- se |x|