| dx | ||
∫ (od −∞ do ∞) | ||
| −x2+2x−3 |
| dx | ||
Umiesz policzyć ∫ | ![]() | |
| −x2+2x−3 |
| 1 | x−1 | |||
Policzyłem i wyszło mi | arctg | . Poźniej mam problemy z | ||
| √2 | √2 |
| (x−1)2 | x−1 | |||
−2*( | +1] = −2*[ ( | )2 + 1] | ||
| 2 | √2 |
| dx | |||||||||||
J = −2∫ | |||||||||||
|
| x−1 | ||
t = | ||
| √2 |
| 1 | ||
dt = | dx | |
| √2 |
| √2dt | dt | x−1 | ||||
J = −2∫ | = −2√2∫ | = −2√2arctgt + C = −2√2arctg | + C | |||
| t2+1 | t2+1 | √2 |
| B−1 | A−1 | |||
Joznaczona =A∫B f(x) = −2√2arctg | + 2√2arctg | |||
| √2 | √2 |
| B−1 | A−1 | |||
Jniewłaściwa = limA→ −∞ [ limB→+∞ −2√2arctg | + 2√2arctg | ] = | ||
| √2 | √2 |
| π | A−1 | |||
limA→ −∞ [ −2√2* | + 2√2arctg | ] = | ||
| 2 | √2 |
| π | π | π | ||||
−2√2* | + 2√2*(− | ) = −4√2* | = −2√2π | |||
| 2 | 2 | 2 |