Let $I=\int \frac{x^{1 / 2}}{1+x^{3 / 4}} d x$

Put $x=t^4 \Rightarrow d x=4 t^3 d t$

$,\therefore\quad I=4 \int \frac{t^2\left(t^3\right)}{1+t^3} d t=4 \int\left(t^2-\frac{t^2}{1+t^3}\right) d t$

$$\begin{aligned} I & =4 \int t^2 d t-4 \int \frac{t^2}{1+t^3} d t \\ I & =I_1-I_2 \\ I_1 & =4 \int t^2 d t=4 \cdot \frac{t^3}{3}+C_1=\frac{4}{3} x^{3 / 4}+C_1 \end{aligned}$$

Now, $$I_2=4 \int \frac{t^2}{1+t^3} d t$$

$$\begin{aligned} &\text { Again, put }\\ &1+t^3=z \Rightarrow 3 t^2 d t=d z\\ &\Rightarrow \quad t^2 d t=\frac{1}{3} d z=\frac{4}{3} \int \frac{1}{z} d z \end{aligned}$$

$$ \begin{aligned} & =\frac{4}{3} \log |z|+C_2=\frac{4}{3} \log \left|\left(1+t^3\right)\right|+C_2 \\ & =\frac{4}{3} \log \left|\left(1+x^{3 / 4}\right)\right|+C_2 \\ \therefore\quad I & =\frac{4}{3} x^{3 / 4}+C_1-\frac{4}{3} \log \left|\left(1+x^{3 / 4}\right)\right|-C_2 \\ & \left.=\frac{4}{3} x^{3 / 4}-\log \right\rvert\,\left(1+x^{3 / 4}\right)+C\quad \left[\because C=C_1-C_2\right] \end{aligned}$$

$$\begin{aligned} \text{Let}\quad I & =\int \frac{\sqrt{1+x^2}}{x^4} d x=\int \frac{\sqrt{1+x^2}}{x} \cdot \frac{1}{x^3} d x \\ & =\int \sqrt{\frac{1+x^2}{x^2}} \cdot \frac{1}{x^3} d x=\int \sqrt{\frac{1}{x^2}+1} \cdot \frac{1}{x^3} d x \end{aligned}$$

$$\begin{aligned} & \text { Put } \quad 1+\frac{1}{x^2}=t^2 \Rightarrow \frac{-2}{x^3} d x=2 t d t \\ & \Rightarrow \quad-\frac{1}{x^3}=t d t \\ & \therefore \quad I=-\int t^2 d t=-\frac{t^3}{3}+C=-\frac{1}{3}\left(1+\frac{1}{x^2}\right)^{3 / 2}+C \end{aligned}$$

Let $I=\int \frac{d x}{\sqrt{16-9 x^2}}=\int \frac{d x}{\sqrt{(4)^2-(3 x)^2}} d x=\frac{1}{3} \sin ^{-1}\left(\frac{3 x}{4}\right)+C$

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

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

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

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