$\int e^x\left(\log x+\frac{1}{x^2}\right)dx$ is equal to: (consider $\log x=\log_e x$)

[C is an arbitrary constant]

The value of $\int_{-4}^{4} \log_e \left( \frac{1-x}{1+x} \right) dx$ is equal to

$\int_{4}^{10} \frac{\log(x^2)}{\log(x^2)+\log(196-28x+x^2)}dx$ is equal to : (Consider $\log_e x = \log x$)

$\int \frac{dx}{e^x + e^{-x}}$ is equal to

If $\displaystyle\int \dfrac{\sin^4 x}{\cos^8 x}\, dx = a\tan^7 x + b\tan^5 x + C$, where 'C' is an arbitrary constant then

If the manufacturer's marginal cost function is $100\left(\frac{1}{\sqrt{x}}+50\right)$, then the cost involved to increase production from 25 units to 100 units is:

$\int (e^{x\log a} + e^{a\log x}) dx$ is equal to (where $a > 1$)