Integrals Test ...

$$\displaystyle \int \dfrac{xe^x}{(1 + x)^2} dx$$ is equal to

Let $$f(x)$$ and $$g(x)$$ be two function satisfying $$f(x^2)+g (4-x)=4x^3, g(4-x)+g(x)=0$$, then the value of $$\int_{-4}^{4} f(x^2)dx$$ is:

If $$\int \dfrac {1}{x \sqrt {1 - x^{3}}}dx = a\log \left |\dfrac {\sqrt {1 - x^{3}} - 1}{\sqrt {1 - x^{3}} + 1}\right | + b$$ then $$a =$$

The value of the definite integral $$\int_{1}^{\infty}(e^{x + 1} + e^{3 - x})^{-1}dx$$ is

Solve $$\int_{0}^{\dfrac {\pi}{2}}\sqrt {\sin \phi}\cos^{5}\phi d\phi$$.

$$\int_{0}^{\infty} \dfrac {x\tan^{-1}x}{(1 + x^{2})x^{2}} dx$$.

The value of $$\displaystyle \int _{ 0 }^{ \dfrac { \pi  }{ 4 }  }{ \cos ^{ 3 }{ 2x } dx } $$ is:

If $$\cfrac { \pi  }{ 2 } <t<\cfrac { 2\pi  }{ 3 } $$ and $$=\int _{ 0 }^{ \sin { 2t }  }{ \cfrac { dx }{ \sqrt { 4\cos ^{ 2 }{ t-{ x }^{ 2 } }  }  }  } $$ then the value of $$\cfrac { 2005\left( I+t \right)  }{ \pi  } $$ equals

The value of $$\int_{0}^{1} \dfrac {8\log (1 + x)}{1 + x^{2}} dx$$ is

The function 
$$ f(x) = \int_1^x [ 2 (t-1) (t-2)^3+3(t-1)^2 (t-2)^2] dt $$ has :