Let $\quad I=\int \frac{x}{x^4-1} d x$

Put $\quad x^2=t \Rightarrow 2 x d x=d t \Rightarrow x d x=\frac{1}{2} d t$

$\therefore \quad I=\frac{1}{2} \int \frac{d t}{t^2-1}=\frac{1}{2} \cdot \frac{1}{2} \log \left|\frac{t-1}{t+1}\right|+C \quad\left[\because \int \frac{d x}{x^2-a^2}=\frac{1}{2 a} \log \left|\frac{x-a}{x+a}\right|+C\right]$

$=\frac{1}{4}\left[\log \left|x^2-1\right|-\log \left|x^2+1\right|\right]+C$

Let $$\begin{aligned} I & =\int \frac{x^2}{1-x^4} d x \\ & =\int \frac{\left(\frac{1}{2}+\frac{x^2}{2}-\frac{1}{2}+\frac{x^2}{2}\right)}{\left(1-x^2\right)\left(1+x^2\right)} d x \end{aligned} \quad\left[\because a^2-b^2=(a+b)(a-b)\right]$$

$$\begin{aligned} & =\int \frac{\frac{1}{2}\left(1+x^2\right)-\frac{1}{2}\left(1-x^2\right)}{\left(1-x^2\right)\left(1+x^2\right)} d x \\ & =\int \frac{\frac{1}{2}\left(1+x^2\right)}{\left(1-x^2\right)\left(1+x^2\right)} d x-\frac{1}{2} \int \frac{\left(1-x^2\right)}{\left(1-x^2\right)\left(1+x^2\right)} d x \\ & =\frac{1}{2} \int \frac{1}{1-x^2} d x-\frac{1}{2} \int \frac{1}{1+x^2} d x=\frac{1}{2} \cdot \frac{1}{2} \log \left|\frac{1+x}{1-x}\right|+C_1-\frac{1}{2} \tan ^{-1} x+C_2 \\ & =\frac{1}{4} \log \left|\frac{1+x}{1-x}\right|-\frac{1}{2} \tan ^{-1} x+C \quad \quad\left[\because C=C_1+C_2\right] \end{aligned}$$

Let $$\begin{aligned} I & =\int \sqrt{2 \mathrm{a} x-x^2} d x=\int \sqrt{-\left(x^2-2 a x\right) d x} \\ & =\int \sqrt{-\left(x^2-2 a x+a^2-a^2\right)} d x=\int \sqrt{-(x-a)^2-a^2} d x \\ & =\int \sqrt{a^2-(x-a)^2} d x \\ & =\frac{x-a}{2} \sqrt{a^2-(x-a)^2}+\frac{a^2}{2} \sin ^{-1}\left(\frac{x-a}{a}\right)+C \\ & =\frac{x-a}{2} \sqrt{2 a x-x^2}+\frac{a^2}{2} \sin ^{-1}\left(\frac{x-a}{a}\right)+C \end{aligned}$$

Let $$I=\int \frac{\sin ^{-1} x}{\left(1-x^2\right)^{3 / 4}} d x=\int \frac{\sin ^{-1} x}{\left(1-x^2\right) \sqrt{1-x^2}} d x$$

Put $\quad \sin ^{-1} x=t \Rightarrow \frac{1}{\sqrt{1-x^2}} d x=d t$

and $$x=\sin t \Rightarrow 1-x^2=\cos ^2 t $$

$\Rightarrow \quad \cos t=\sqrt{1-x^2}$

$\therefore \quad I=\int \frac{t}{\cos ^2 t} d t=\int t \cdot \sec ^2 t d t$

$$\begin{aligned} & =t \cdot \int \sec ^2 t d t-\int\left(\frac{d}{d t} t \cdot \int \sec ^2 t d t\right) d t \\ & =t \cdot \tan t-\int 1 \cdot \tan t d t \\ & =t \tan t+\log |\cos t|+C \quad \quad\left[\because \int \tan x d x=-\log |\cos x|+C\right] \\ & =\sin ^{-1} x \cdot \frac{x}{\sqrt{1-x^2}}+\log \left|\sqrt{1-x^2}\right|+C \end{aligned}$$

Let $I=\int \frac{\cos 5 x+\cos 4 x}{1-2 \cos 3 x} d x=\int \frac{2 \cos \frac{9 x}{2} \cdot \cos \frac{x}{2}}{1-2\left(2 \cos ^2 \frac{3 x}{2}-1\right)} d x$

$\left[\because \cos C+\cos D=2 \cos \frac{C+D}{2} \cdot \cos \frac{C-D}{2}\right.$ and $\left.\cos 2 x=2 \cos ^2 x-1\right]$

$I=\int \frac{2 \cos \frac{9 x}{2} \cdot \cos \frac{x}{2}}{3-4 \cos ^2 \frac{3 x}{2}} d x=-\int \frac{2 \cos \frac{9 x}{2} \cdot \cos \frac{x}{2}}{4 \cos ^2 \frac{3 x}{2}-3} d x$

$=-\int \frac{2 \cos \frac{9 x}{2} \cdot \cos \frac{x}{2} \cdot \cos \frac{3 x}{2}}{4 \cos ^3 \frac{3 x}{2}-3 \cos \frac{3 x}{2}} d x \quad\left[\right.$ multiply and divide by $\left.\cos \frac{3 x}{2}\right]$

$$ \begin{aligned} & =-\int \frac{2 \cos \frac{9 x}{2} \cdot \cos \frac{x}{2} \cdot \cos \frac{3 x}{2}}{\cos 3 \cdot \frac{3 x}{2}} d x=-\int 2 \cos \frac{3 x}{2} \cdot \cos \frac{x}{2} d x \\ & =-\int\left\{\cos \left(\frac{3 x}{2}+\frac{x}{2}\right)+\cos \left(\frac{3 x}{2}-\frac{x}{2}\right)\right\} d x \\ & =-\int(\cos 2 x+\cos x) d x \\ & =-\left[\frac{\sin 2 x}{2}+\sin x\right]+C \\ & =-\frac{1}{2} \sin 2 x-\sin x+C \end{aligned}$$