Całka oznaczona

[ullenka88]

1) \(\displaystyle{ \int \frac{1}{\sqrt{ -x^2+3x-3}}}\)
2) \(\displaystyle{ \int \limits_{-\infty}^{0} e^-^1^0dx}\)
3) \(\displaystyle{ \int\limits_{-1}^{1}\frac{1}{e^6^x-1}dx}\)
4) \(\displaystyle{ \int \frac {1}{cosx}dx}\)

[skowron]

Pierwsza:
\(\displaystyle{ \ldots = t \frac{1}{\sqrt{-\frac{3}{4}-(x-\frac{3}{2})^{2}}} = ln|x-\frac{3}{2}+\sqrt{-\frac{3}{4}-(x-\frac{3}{2})^{2}}|+C}\)

[Szemek]

3) nieoznaczona:
\(\displaystyle{ \int \frac{dx}{e^{6x}-1} = ft|\begin{array}{l} t=e^{6x} \\ dt=6e^{6x}dx \\ \frac{1}{6}dt=e^{6x}dx \end{array}\right| = \frac{1}{6} t \frac{dt}{t(t-1)} = \frac{1}{6}\left( t \frac{dt}{t-1} - t \frac{dt}{t} \right) = \\ = \frac{1}{6}\left( \ln|t-1| - \ln|t|\right)+C = \frac{1}{6}\ln|e^{6x}-1|-\frac{1}{6}\ln|e^{6x}|+C =\underline{ \frac{1}{6}\ln|e^{6x}-1|-x+C}}\)

[skowron]

Druga:
\(\displaystyle{ \ldots = e^{-10} \lim_{a \to } [\left x ]\right_{-a}^{0} = }\)
Czy mógł by ktoś pod twierdzić to rozwiązanie?