$$\int {{{\cos 2x} \over {{{\cos }^2}x{{\sin }^2}x}}dx = } $$
Solve $$\displaystyle\int \dfrac{x}{{{{\left( {x + 1} \right)}^2}}}dx$$
Find $$\int a^{x} \cdot e^{x} dx$$.
$$\int {\dfrac{1}{{{e^x} + 1}}} dx$$
value of $$\int _{ -2 }^{ 3 }{ \left| 1-x^{ 2 } \right| dx= } $$