1 | ||
2x = t4 − 3 ⇔ x = | (t4 − 3) | |
2 |
1 | ||
∫x*(2x + 3)14dx = ∫ | (t4 − 3)t*2t3dt = ∫(t4 − 3)t4dt = ∫t16dt − 3∫t4dt = | |
2 |
1 | 3 | 1 | 3 | |||||
= | t17 − | t5 + C = | (t4)4t − | t4t + C = | ||||
17 | 5 | 17 | 5 |
1 | 3 | |||
= | (2x + 3)4 * (2x + 3)14 − | ((2x + 3) * (2x + 3)14) + C | ||
17 | 5 |
5^2 | 52 |
2^{10} | 210 |
a_2 | a2 |
a_{25} | a25 |
p{2} | √2 |
p{81} | √81 |
Kliknij po więcej przykładów | |
---|---|
Twój nick | |