\( \left\{ \begin{array} { l } { f ( x ) = \frac { \ln x - 1 } { \ln x } ; x > 0 } \\ { f ( 0 ) = 0 } \end{array} \right. \)

If the slope of the tangent to the curve: \( y=f(x) \) at any point on it equals \( 6 x+a \), whe \( a \) is constant if the equation of the tangent to the curve at the point \( (1,-1) \) is \( y=4- \) \( 5 x \), then the equation of the curve is ... \( \begin{array}{llll}\text { a) } y=3 x^{2}-4 & \text { b) } y=3 x^{2}-11 x+7 & \text { c) } y=3 x^{2}-4 x & \text { d) } y=3 x^{2}-x-3\end{array} \)

If the slope of the tangent to the curve: \( y=f(x) \) at any point on it equals \( 6 x+a \), where \( a \) is constant if the equation of the tangent to the curve at the point \( (1,-1) \) is \( y=4- \) \( 5 x \), then the equation of the curve is ... \( \begin{array}{llll}\text { a) } y=3 x^{2}-4 & \text { b) } y=3 x^{2}-11 x+7 & \text { c) } y=3 x^{2}-4 x & \text { d) } y=3 x^{2}-x-3\end{array} \)

Determine the indefinite integral \( \int (6x^3 + 4x^2) \, dx \)

3. \( \lim _{x \rightarrow \infty} \frac{x-1}{2 x+5} \)

nction définie par \( f(x)=\frac{L}{x^{2}+1} \)