Embibe Experts Solutions for Exercise 1: Assignment

Let $\underset{n\to \infty }{\lim }{\left(\frac{\left(kn\right)!}{n^{kn}}\right)}^{\frac{1}{n}}=\lambda \left(k\right)$, where $k\in N$

${\int }_0^{2\pi }\frac{x}{1+{\tan }^{24}x}dx, n>0$

Let $I={\int }_0^n\left[x\right]dx, n>0$, where $\left[\right]$ is G.I.F.,

If $f\left(x\right)={\int }_0^{{\cos }^{2}x}\sec x\frac{\sqrt{t}}{1+t^3}dt$, then

Let $I_{n, m}=\int \frac{{\sin }^{n}x}{{\cos }^{m}x}dx$. Which of the following options is/are correct?

If $\left[.\right]$ denotes the greatest integer function and $\left\{x\right\}=x-\left[x\right]$, then the value of ${\int }_0^{1}x\left\{\frac{1}{x}\right\}\left[\frac{1}{x}\right]dx$ is equal to, if $\sum _{k=1}^{\infty }\frac{1}{k^2}=\frac{{\pi }^2}{6}$

${\int }_1^2\left(\ln \left(ex\right)\right)x^{x}dx$ is equal to

${\int }_0^{\frac{1}{\sqrt{3}}}\sqrt{x+\sqrt{x^2+1}}dx$ is equal to