Integrals Test ...

Let $$\displaystyle I_1 = \int_1^2 \frac{1}{\sqrt{1 + x^2}} dx$$ and $$I_2 \displaystyle = \int_1^2 \frac{1}{x} dx$$. Then

The value of the definite integral $$\displaystyle \int_{ 0 }^{\sqrt{{\ln (\displaystyle \frac{\pi}{2})}}}\cos(e^{x^{2}})2xe^{x^{2}} \:dx$$ is:

The evaluation of $$\displaystyle \int \frac{pX^{p+2q-1}-qX^{q-1}}{X^{2p+2q}+2X^{p+q}+1}dx$$ is 

$$\displaystyle \int_{1}^{e^{37}}\frac{\pi \sin \left ( \pi \log _{e}x \right )}{x}dx$$ is equal to

$$\displaystyle \int_{0}^{2}\sqrt{\frac{2+x}{2-x}}dx$$ is equal to

$$\displaystyle\int_0^\infty{f\left(x+\frac{1}{x}\right).\frac{\ln{x}}{x}dx}$$ is equal to

Evaluate $$\displaystyle\int_0^{\displaystyle\frac{\pi}{4}}{\frac{\sin{x}+\cos{x}}{9+16\{1-{(\sin{x}-\cos{x})}^2\}}dx}$$

If $$\displaystyle 2f(x)+f(-x)=\frac{1}{x}\sin{\left(x-\frac{1}{x}\right)}$$, then the value of $$\displaystyle\int_{\frac{1}{e}}^{e}{f(x)dx}$$, is

If $$x$$ satisfies the equation $$\displaystyle\left(\int_0^1{\frac{dt}{t^2+2t\cos{\alpha}+1}}\right)x^2-\left(\int_{-3}^3{\frac{t^2\sin{2t}}{t^2+1}dt}\right)x-2=0$$ 

The tangent to the graph of the function $$\displaystyle y = f(x)$$ at the point with abscissa x = a forms with the x-axis an angle of $$\displaystyle \pi/3$$ and at the point with abscissa x = b at an angle of $$\displaystyle \pi/4$$, then the value of the integral,
$$\displaystyle \int_{a}^{b} f'(x).f''(x)dx$$ is equal to