$\int \frac{1}{x(x^6+1)}\,dx =$ (consider $\log_e x = \log x$)

$\int \frac{x^2 + 4}{(x^2 + 3)(x^2 + 5)} \, dx = u \tan^{-1}\left(\frac{x}{\sqrt{3}}\right) + v \tan^{-1}\left(\frac{x}{\sqrt{5}}\right) + c$, where c is arbitrary constant, then value of $\frac{1}{u^2} + \frac{1}{v^2}$ is equal to :

$\int e^{2x}(\sin x + \frac{1}{2}\cos x) dx$ is equal to

In which of the following cases the value of the integral is more than 10?

A. $\displaystyle\int_0^2 |x|\,dx$

B. $\displaystyle\int_0^2 (x+3)^2\,dx$

C. $\displaystyle\int_0^2 e^x\,dx$

D. $\displaystyle\int_0^2 (2x+5)\,dx$

Choose the correct answer from the options given below:

$\int_{-\pi}^{\pi} \frac{e^{\sin x}}{e^{\sin x} + e^{-\sin x}}dx$ is equal to

$\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{dx}{1 + \sqrt{tanx}}$ is equal to

$\int_0^1 \frac{dx}{\sqrt{1+x} - \sqrt{x}}$ is equal to