1-1-3x-4-3-4x-3x-2-dt-please-help-me-

Question Number 152588 by rexford last updated on 30/Aug/21

\int_{- 1}^{1} \frac{3 x + 4}{3 + 4 x + 3 x^{2}} d t

p l e a s e, h e l p m e

Answered by qaz last updated on 30/Aug/21

\int_{- 1}^{+ 1} \frac{3 x + 4}{3 + 4 x + {3 x}^{2}} dt = \frac{3 x + 4}{3 + 4 x + {3 x}^{2}} \cdot t ∣_{- 1}^{1} = \frac{6 x + 8}{3 + 4 x + {3 x}^{2}}

Answered by puissant last updated on 30/Aug/21

= \frac{1}{2} \int_{- 1}^{1} \frac{6 x + 4 + 4}{3 x^{2} + 4 x + 3} d x

= \frac{1}{2} \int_{- 1}^{1} \frac{6 x + 4}{3 x^{2} + 4 x + 3} d x + \frac{2}{3} \int_{- 1}^{+ 1} \frac{1}{x^{2} + \frac{4}{3} x + 1} d x

= \frac{1}{2} {[l n ∣ 3 x^{2} + 4 x + 3 ∣]}_{- 1}^{1} + \frac{2}{3} Q

= \frac{1}{2} l n 5 + \frac{2}{3} Q

Q = \int_{- 1}^{1} \frac{1}{x^{2} + \frac{4}{3} x + 1} d x = \int_{- 1}^{1} \frac{1}{{(x + \frac{2}{3})}^{2} - \frac{4}{9} + \frac{9}{9}} d x

= \int_{- 1}^{1} \frac{1}{{(x + \frac{2}{3})}^{2} + \frac{5}{9}} d x = \frac{9}{5} \int_{- 1}^{1} \frac{1}{[{(\frac{3}{\sqrt{5}} (x + \frac{2}{3}))}^{2} + 1]} d x

u = \frac{3}{\sqrt{5}} (x + \frac{2}{3}) \to d x = \frac{\sqrt{5}}{3} d u

\Rightarrow Q = \frac{9}{5} \times \frac{\sqrt{5}}{3} \int_{- 1}^{1} \frac{1}{u^{2} + 1} d u

= \frac{3}{\sqrt{5}} {[a r c t a n (u)]}_{- 1}^{1} = \frac{3}{\sqrt{5}} \times \frac{π}{2} .

∴∵ I = \frac{1}{2} l n 5 + \frac{3}{\sqrt{5}} \times \frac{2}{3} \times \frac{π}{2}

= \frac{1}{2} l n 5 + \frac{π}{\sqrt{5}} . .

Answered by phanphuoc last updated on 30/Aug/21

= 1 / 2 \int_{- 1}^{1} \frac{6 x + 4 d x}{3 + 4 x + 3 x^{2}} + 2 \int_{- 1}^{1} \frac{d x}{3 + 4 x + 3 x^{2}}

= 1 / 2 l n {(1 + 4 x + 3 x^{2})}_{- 1}^{1} + 2 / 3 \int_{- 1}^{1} \frac{d x}{1 + 4 / 3 x + x^{2}}

= \dots + 2 / \sqrt{5} a r c t a n {(\frac{x + 2 / 3}{\sqrt{5} / 3})}_{- 1}^{1} ==== \dots . .