\(\int \frac{\sec ^{2} x}{\tan ^{2} x-1} d x =\int \frac{1}{(\tan x)^{2}-1} \times \sec ^{2} x d x \)
\(=\int \frac{1}{u^{2}-1} d u, \quad u=\tan x \)
\(=\frac{1}{2(1)} \log \left|\frac{u-1}{u+1}\right|+c \)
\(\frac{1}{2} \log \left|\frac{\tan x-1}{\tan x+1}\right|+c \)

\( p v=81\)
\( p=\frac{81}{v}=\frac{81}{9}=9 \)
Diff w. r. to v
\( p(1)+v \cdot \frac{d p}{d v} =0\)
\(v \frac{d p}{d v} =-p \)
\(\frac{d p}{d v} =\frac{-p}{v}=\frac{-9}{9}=-1\)
\(\frac{d p}{d v} =-1\)

\(\lim _{n \rightarrow \infty}[\left.\frac{1}{n^{2}}+\frac{2}{n^{2}}+\frac{3}{n^{2}}+\cdots+\frac{n}{n^{2}}\right] \)
\(=\lim _{n \rightarrow \infty}\left[\frac{(1+2+3+\cdots+\dot{n})}{n^{2}}\right] \)
\(=\lim _{n \rightarrow \infty}\left[\frac{n(n+1)}{2 n^{2}}\right] \)
\(=\frac{1}{2} \lim _{n \rightarrow \infty}\left(\frac{n+1}{n}\right) \)
\(=\frac{1}{2} \lim _{n \rightarrow \infty}\left(1+\frac{1}{n}\right)=1 / 2(1+0)=1 / 2 \)

\(n^{\text {th }} \text { term }=1-\frac{1}{2^{n}}=1-2^{-n}\)

\(f(x) =\frac{\left(x^{2}+\cos x\right)\left(1+x^{4}\right)}{(x-\sin x)\left(2 x-x^{3}\right)}+e^{-|x|} \)
\(f(-x) =\frac{\left[(-x)^{2}+\cos (-x)\right]\left[1+(-x)^{4}\right]}{[-x-\sin (-x)]\left[-2 x-\left(-x^{3}\right)\right]}+e^{+x \mid} \)
\(=\frac{\left(x^{2}+\cos x\right)\left(1+x^{4}\right)}{(x-\sin x)\left(2 x-x^{3}\right)}+e^{-|x|} \)
\(=f(x) \)
Hence f(x) is an even function