Integrals Test ...

Let $$\displaystyle\frac{d}{dx}(F(x))=\frac{e^{\displaystyle\sin{x}}}{x}$$, $$x>0$$. If $$\displaystyle\int_1^4{\frac{2e^{\displaystyle\sin{x^2}}}{x}dx}=F(k)-F(1)$$, then the possible value of $$k$$ is

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

$$\displaystyle \int_{0}^{\infty}\frac{1}{1+x^{n}}dx,\:\forall\:n\:> 1$$ is equal to?

The value of $$\displaystyle \int_{1}^{1/e}f(x)dx+\int_{1}^{e}f(x)dx$$ where $$f(x)$$ is given as $$\displaystyle \frac{log\:x}{1+x}$$ equals 

If $$\displaystyle I= \int_{1/\pi }^{\pi }\frac{1}{x}\cdot \sin \left ( x-\frac{1}{x} \right )dx$$ then I is equal to

The value (s) of  $$\displaystyle \int_{0}^{1}\frac{x^{4}\left ( 1-x \right )^{4}}{1+x^{2}} dx $$ is (are)

If $$\displaystyle I= \int_{0}^{1}\frac{x dx}{8+x^{3}}$$ then the smallest interval in which $$I$$ lies is

$$ \displaystyle \int_{1}^{\infty }\frac{\log \left ( t-1 \right )}{t^2\log t+\log \left ( \frac{t}{t-1} \right )}\:dt $$ equals

The value of $$ \displaystyle \int_{0}^{\pi /2}\sin \theta \log \left ( \sin \theta \right )\:d\theta  $$ equals

If $$\displaystyle I_{n} =\int_{0}^{\frac{\pi }{4}}\tan ^{n}xdx$$