Integrals Test ...

The value of integral  $$\displaystyle \int_{0}^{\infty }\frac{x\log x}{(1+x^2)^2}  \: dx$$ is

Let $$\displaystyle \frac{{df(x)}}{{dx}} = \frac{{{e^{\sin x}}}}{x}, x>0$$. If $$\displaystyle \int_1^4 {\frac{{3{e^{\sin {x^3}}}}}{x}dx = f(k) - f(1)} $$ then one of the possible values of k is

Let $$f(0) = 0$$ and $$\displaystyle \int_{0}^{2}{f}'(2t)e^{f(2t)} \:dt=5$$.

$$\displaystyle \int_{-3\pi/2}^{-\pi/2}[(x+\pi)^{3}+\cos^{2}(x+3\pi)]dx=$$

The value of the integral $$\displaystyle \int_{0}^{1}\frac{1}{(x^{2}+1)^{3/2}} dx$$ is

If $$\displaystyle \int^{x}_{\log 2} \dfrac{1}{\sqrt{e^{x}-1}}dx = \dfrac{\pi}{6}$$

Let $$\displaystyle \frac{d}{dx}F\left ( x \right )=\frac{e^{\sin x}}{x},x> 0.$$ If $$\displaystyle \int_{1}^{4}\frac{2e^{\sin x^{2}}}{x}dx=F\left ( k \right )-F\left ( 1 \right )$$ then one of the possible values of $$\displaystyle k$$ is

If $$\displaystyle \int_{0}^{1}xe^{x^{2}}dx= \lambda \int_{0}^{1}e^{x^{2}}dx$$ then $$\lambda $$

The value of $$\displaystyle \int_{1}^{2}\left [ f\left \{ g\left ( x \right ) \right \} \right ]^{-1}.{f}'\left \{ g\left ( x \right ) \right \}.{g}'\left ( x \right )dx,$$ where $$\displaystyle g\left ( 1 \right )=g\left ( 2 \right ),$$ is equal to?

The value of  $$\displaystyle \int_{0}^{\pi /4}\frac{\sec x}{\left ( \sec x+\tan x \right )^{2}}dx$$ is