Calcule as seguintes integrais indefinidas por frações parciais:
(a)
integral 1x^2-1 d x
(b)
integral -2x^2-16 d x
(c)
integral 13 x^2-x d x
(d)
integral 12 x^2+x d x
(e)
integral 3x^2+x-2 d x
(f)
integral 5-x2 x^2+x-1 d x
(g)
integral x^2+12 x+12x^3-4 x d x
(h)
integral x+2x^2-4 x d x
(i)
integral 4-3 x(x-1)^2 d x
(j)
integral 4 x^2-12 x(x^2+2 x+1) d x

Passo 1

E aí, vamos calcular dez integrais indefinidas usando o método de frações parciais. A ideia é:
  1. Fatorar o denominador Q(x).
  2. Escrever P(x)/Q(x) como soma de termos simples do tipo A/(x–a) ou A/(x–a)ⁿ.
  3. Determinar as constantes A, B, C… resolvendo um sisteminha.
  4. Integrar cada pedaço usando ∫1/u du=ln|u| ou ∫uᵐ du=uᵐ⁺¹/(m+1).
    Bora nessa!

Passo 2

Resolvendo (a) ∫1/(x²−1) dx
  1. Fatora: x²−1=(x−1)(x+1).
  2. Supõe 1/(x²−1)=A/(x−1)+B/(x+1).
  3. Multiplica: 1=A(x+1)+B(x−1).
  4. Põe x=1 ⇒ 1=2A ⇒ A=1/2.
    Põe x=−1 ⇒ 1=−2B ⇒ B=−1/2.
  5. ∫1/(x²−1)dx=∫(½·1/(x−1) −½·1/(x+1))dx
    =½ ln|x−1| −½ ln|x+1| +C
    =½ ln| (x−1)/(x+1) | +C

Passo 3

Resolvendo (b) ∫−2/(x²−16) dx
  1. x²−16=(x−4)(x+4).
  2. −2/(…)=A/(x−4)+B/(x+4).
  3. −2=A(x+4)+B(x−4).
  4. x=4 ⇒ −2=8A ⇒ A=−1/4.
    x=−4 ⇒ −2=−8B ⇒ B=1/4.
  5. ∫−2/(x²−16)dx=∫(−¼·1/(x−4)+¼·1/(x+4))dx
    =−¼ ln|x−4|+¼ ln|x+4|+C

Passo 4

Resolvendo (c) ∫1/(3x²−x) dx
  1. 3x²−x=x(3x−1).
  2. 1/(x(3x−1))=A/x+B/(3x−1).
  3. 1=A(3x−1)+Bx ⇒ (3A+B)x−A.
  4. −A=1⇒A=−1; 3A+B=0⇒B=3.
  5. ∫(−1/x+3/(3x−1))dx=−ln|x|+3·(1/3)ln|3x−1|+C
    =−ln|x|+ln|3x−1|+C

Passo 5

Resolvendo (d) ∫1/(2x²+x) dx
  1. 2x²+x=x(2x+1).
  2. 1/[x(2x+1)]=A/x+B/(2x+1).
  3. 1=A(2x+1)+Bx ⇒ (2A+B)x + A.
  4. A=1⇒2·1+B=0⇒B=−2.
  5. ∫(1/x−2/(2x+1))dx=ln|x|−2·(1/2)ln|2x+1|+C
    =ln|x|−ln|2x+1|+C

Passo 6

Resolvendo (e) ∫3/(x²+x−2) dx
  1. x²+x−2=(x+2)(x−1).
  2. 3/(…)=A/(x+2)+B/(x−1).
  3. 3=A(x−1)+B(x+2) ⇒ (A+B)x + (−A+2B).
  4. A+B=0; −A+2B=3 ⇒ B=1, A=−1.
  5. ∫(−1/(x+2)+1/(x−1))dx=−ln|x+2|+ln|x−1|+C

Passo 7

Resolvendo (f) ∫(5−x)/(2x²+x−1) dx
  1. 2x²+x−1=(2x−1)(x+1).
  2. (5−x)/…=A/(2x−1)+B/(x+1).
  3. 5−x=A(x+1)+B(2x−1) ⇒ (A+2B)x + (A−B).
  4. A+2B=−1; A−B=5 ⇒ A=3, B=−2.
  5. ∫(3/(2x−1)−2/(x+1))dx=3·½ln|2x−1|−2ln|x+1|+C
    =3/2 ln|2x−1|−2 ln|x+1|+C

Passo 8

Resolvendo (g) ∫(x²+12x+12)/(x³−4x) dx
  1. x³−4x = x(x−2)(x+2).
  2. Decompõe =A/x +B/(x−2)+C/(x+2).
  3. x²+12x+12 =A(x−2)(x+2)+B x(x+2)+C x(x−2).
    Compara: A+B+C=1; 2B−2C=12; −4A=12⇒A=−3;
    B−C=6 e B+C=4 ⇒ B=5, C=−1.
  4. ∫(−3/x+5/(x−2)−1/(x+2))dx
    =−3ln|x|+5ln|x−2|−ln|x+2|+C

Passo 9

Resolvendo (h) ∫(x+2)/(x²−4x)dx
  1. x²−4x=x(x−4).
  2. =A/x+B/(x−4).
  3. x+2=A(x−4)+Bx ⇒ (A+B)x−4A.
  4. A+B=1; −4A=2⇒A=−1/2; B=3/2.
  5. ∫(−½·1/x+3/2·1/(x−4))dx
    =−½ln|x|+3/2ln|x−4|+C

Passo 10

Resolvendo (i) ∫(4−3x)/(x−1)² dx
  1. Uso A/(x−1)+B/(x−1)².
  2. 4−3x=A(x−1)+B ⇒ Ax + (B−A).
  3. A=−3; B−A=4⇒B=1.
  4. ∫(−3/(x−1)+1/(x−1)²)dx
    =−3ln|x−1| + ∫(x−1)⁻²dx
    =−3ln|x−1| −1/(x−1) +C

Passo 11

Resolvendo (j) ∫(4x²−1)/[2x(x+1)²]dx
  1. Puxa 1/2 e fatora x(x+1)².
  2. Decompõe em A/x +B/(x+1)+C/(x+1)².
  3. 4x²−1=A(x+1)²+B x(x+1)+C x.
    Compara: A+B=4; 2A+B+C=0; A=−1 ⇒ B=5, C=−3.
  4. ∫=½∫(−1/x+5/(x+1)−3/(x+1)²)dx
    =½[−ln|x|+5ln|x+1|+3/(x+1)] +C
    =−½ln|x|+5/2ln|x+1|+3/[2(x+1)] +C

Resposta

(a) ½ ln| (x−1)/(x+1) |+C
(b) −¼ ln|x−4|+¼ ln|x+4|+C
(c) −ln|x|+ln|3x−1|+C
(d) ln|x|−ln|2x+1|+C
(e) −ln|x+2|+ln|x−1|+C
(f) 3/2 ln|2x−1|−2 ln|x+1|+C
(g) −3 ln|x|+5 ln|x−2|−ln|x+2|+C
(h) −½ ln|x|+3/2 ln|x−4|+C
(i) −3 ln|x−1|−1/(x−1)+C
(j) −½ ln|x|+5/2 ln|x+1|+3/[2(x+1)]+C

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