$$\begin{aligned} \text { Let }\quad &I=\int \frac{x}{\sqrt{x}+1} d x\\ &\text { Put } \quad \sqrt{x}=t \Rightarrow \frac{1}{2 \sqrt{x}} d x=d t \end{aligned}$$

$$\begin{array}{lrl} \Rightarrow & d x & =2 \sqrt{x} d t \\ \therefore & I & =2 \int\left(\frac{x \sqrt{x}}{t+1}\right) d t=2 \int \frac{t^2 \cdot t}{t+1} d t=2 \int \frac{t^3}{t+1} d t \end{array}$$

$$\begin{aligned} & =2 \int \frac{t^3+1-1}{t+1} d t=2 \int \frac{(t+1)\left(t^2-t+1\right)}{t+1} d t-2 \int \frac{1}{t+1} d t \\ & =2 \int\left(t^2-t+1\right) d t-2 \int \frac{1}{t+1} d t \\ & =2\left[\frac{t^3}{3}-\frac{t^2}{2}+t-\log |(t+1)|\right]+C \\ & =2\left[\frac{x \sqrt{x}}{3}-\frac{x}{2}+\sqrt{x}-\log |(\sqrt{x}+1)|\right]+C \end{aligned}$$

Let $\quad I=\int \sqrt{\frac{a+x}{a-x}} d x$

Put $$x=\operatorname{acos} 2 \theta$$

$$\begin{array}{ll} \Rightarrow & d x=-a \cdot \sin 2 \theta \cdot 2 \cdot d \theta \\ \therefore & I=-2 \int \sqrt{\frac{a+a \cos 2 \theta}{a-a \cos 2 \theta}} \cdot a \sin 2 \theta d \theta \end{array}$$

$\left[\because \cos 2 \theta=\frac{x}{a} \Rightarrow 2 \theta=\cos ^{-1} \frac{x}{a} \Rightarrow \theta=\frac{1}{2} \cos ^{-1} \frac{x}{a}\right]$

$$\begin{aligned} & =-2 a \int \sqrt{\frac{1+\cos 2 \theta}{1-\cos 2 \theta}} \sin 2 \theta d \theta=-2 a \int \sqrt{\frac{2 \cos ^2 \theta}{2 \sin ^2 \theta}} \sin 2 \theta d \theta \\ & =-2 a \int \cot \theta \cdot \sin 2 \theta d \theta=-2 a \int \frac{\cos \theta}{\sin \theta} \cdot 2 \sin \theta \cdot \cos \theta d \theta \\ & =-4 a \int \cos ^2 \theta d \theta=-2 a \int(1+\cos 2 \theta) d \theta \\ & =-2 a\left[\theta+\frac{\sin 2 \theta}{2}\right]+C \\ & =-2 a\left[\frac{1}{2} \cos ^{-1} \frac{x}{a}+\frac{1}{2} \sqrt{1-\frac{x^2}{a^2}}\right]+C \\ & =-a\left[\cos ^{-1}\left(\frac{x}{a}\right)+\sqrt{1-\frac{x^2}{a^2}}\right]+C \end{aligned}$$

Alternate Method

Let $$I=\int \sqrt{\frac{a+x}{a-x}} d x=\int \sqrt{\frac{(a+x)(a+x)}{(a-x)(a+x)}} d x$$

$$\begin{aligned} & =\int \frac{(a+x)}{\sqrt{a^2-x^2}} d x \\ I & =\int \frac{a}{\sqrt{a^2-x^2}}+\int \frac{x}{\sqrt{a^2-x^2}} d x \end{aligned}$$

$\therefore \quad I=I_1+I_2\quad\text{..... (i)}$

Now, $$I_1=\int \frac{a}{\sqrt{a^2-x^2}}=a \sin ^{-1}\left(\frac{x}{a}\right)+C_1$$

and $$I_2=\int \frac{x}{\sqrt{a^2-x^2}} d x$$

Put $\quad a^2-x^2=t^2 \Rightarrow-2 x d x=2 t d t$

$\therefore \quad I_2=-\int \frac{t}{t} d t e=-\int 1 d t$

$$\begin{array}{rlrl} & =-t+C_2=-\sqrt{a^2-x^2}+C_2 & \\ \therefore\quad I & =a \sin ^{-1}\left(\frac{x}{a}\right)+C_1-\sqrt{a^2-x^2}+C_2 & & {\left[\because t^2=a^2-x^2\right]} \\ I & =a^{\sin ^{-1}}\left(\frac{x}{a}\right)-\sqrt{a^2-x^2}+C & & {\left[\because C=C_1+C_2\right]} \end{array}$$

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