Let $$\begin{aligned} I & =\int \frac{d t}{\sqrt{3 t-2 t^2}}=\frac{1}{\sqrt{2}} \int \frac{d t}{\sqrt{-\left(t^2-\frac{3}{2} t\right)}} \\ & =\frac{1}{\sqrt{2}} \int \frac{d t}{\sqrt{-\left[\left(t^2-2 \cdot \frac{1}{2} \cdot \frac{3}{2} t\right)+\left(\frac{3}{4}\right)^2-\left(\frac{3}{4}\right)^2\right]}} \\ & =\frac{1}{\sqrt{2}} \int \frac{d t}{\sqrt{-\left[\left(t-\frac{3}{4}\right)^2-\left(\frac{3}{4}\right)^2\right]}} \\ & =\frac{1}{\sqrt{2}} \int \frac{d t}{\sqrt{\left(\frac{3}{4}\right)^2-\left(t-\frac{3}{4}\right)^2}} \\ & =\frac{1}{\sqrt{2}} \sin ^{-1}\left(\frac{t-\frac{3}{4}}{\frac{3}{4}}\right)+C=\frac{1}{\sqrt{2}} \sin ^{-1}\left(\frac{4 t-3}{3}\right)+C \end{aligned}$$

Let $$\begin{aligned} & I=\int \frac{3 x-1}{\sqrt{x^2+9}} d x \\ & I=\int \frac{3 x}{\sqrt{x^2+9}} d x-\int \frac{1}{\sqrt{x^2+9}} d x \\ & I=I_1-I_2 \end{aligned}$$

Now, $$I_1=\int \frac{3 x}{\sqrt{x^2+9}}$$

Put $$x^2+9=t^2 \Rightarrow 2 x d x=2 t d t \Rightarrow x d x=t d t$$

$\therefore \quad I_1=3 \int_t^t d t$

$=3 \int d t=3 t+C_1=3 \sqrt{x^2+9}+C_1$

$$\begin{aligned} \text{and}\quad I_2 & =\int \frac{1}{\sqrt{x^2+9}} d x=\int \frac{1}{\sqrt{x^2+(3)^2}} d x \\ & =\log \left|x+\sqrt{x^2+9}\right|+C_2 \\ \therefore\quad I & =3 \sqrt{x^2+9}+C_1-\log \left|x+\sqrt{x^2+9}\right|-C_2 \\ & =3 \sqrt{x^2+9}-\log \left|x+\sqrt{x^2+9}\right|+C\quad \left[\because C=C_1-C_2\right] \end{aligned}$$

Let $$\begin{aligned} I & =\int \sqrt{5-2 x+x^2} d x=\int \sqrt{x^2-2 x+1+4} d x \\ & =\int \sqrt{(x-1)^2+(2)^2} d x=\int \sqrt{(2)^2+(x-1)^2} d x \\ & =\frac{x-1}{2} \sqrt{2^2+(x-1)^2}+2 \log \left|x-1+\sqrt{2^2+(x-1)^2}\right|+C \\ & =\frac{x-1}{2} \sqrt{5-2 x+x^2}+2 \log \left|x-1+\sqrt{5-2 x+x^2}\right|+C \end{aligned}$$

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

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