$$\begin{aligned} \text{Let}\quad I & =\int \frac{2 x-1}{2 x+3} d x=\int \frac{2 x+3-3-1}{2 x+3} d x \\ & =\int 1 d x-4 \int \frac{1}{2 x+3} d x=x-\int \frac{4}{2\left(x+\frac{3}{2}\right)} d x \\ & =x-2 \log +\left|\left(x+\frac{3}{2}\right)\right| C^{\prime}=x-2 \log \left|\left(\frac{2 x+3}{2}\right)\right|+C^{\prime} \end{aligned}$$

$$\begin{array}{rlr} =x-2 \log |(2 x+3)|+2 \log 2+C^{\prime} & {\left[\because \log \frac{m}{n}=\log m-\log n\right]} \\ =x-\log \left|(2 x+3)^2\right|+C & {\left[\because C=2 \log 2+C^{\prime}\right]} \end{array}$$

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

$$\begin{array}{lc} \text { Put } & x^2+3 x=t \\ \Rightarrow & (2 x+3) d x=d t \end{array}$$

$$\begin{aligned} \therefore\quad I & =\int \frac{1}{t} d t=\log |t|+C \\ & =\log \left|\left(x^2+3 x\right)\right|+C \end{aligned}$$

Let $I=\int \frac{\left(x^2+2\right) d}{x+1} x$

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

$$\begin{aligned} \text{Let}\quad I & =\int\left(\frac{e^{6 \log x}-e^{5 \log x}}{e^{4 \log x}-e^{3 \log x}}\right) d x \\ & =\int\left(\frac{e^{\log x^6}-e^{\log x^5}}{e^{\log x^4}-e^{\log x^3}}\right) d x \quad \left[\because \operatorname{alog} b=\log b^a\right]\\ & =\int\left(\frac{x^6-x^5}{x^4-x^3}\right) d x \quad \left[\because e^{\log x}=x\right]\\ & =\int\left(\frac{x^3-x^2}{x-1}\right) d x=\int \frac{x^2(x-1)}{x-1} d x \\ & =\int x^2 d x=\frac{x^3}{3}+C \end{aligned}$$

Consider that, $$I=\int \frac{(1+\cos x)}{(x+\sin x)} d x$$

Let $$x+\sin x=t \Rightarrow(1+\cos x) d x=d t$$

$$\begin{aligned} \therefore\quad I & =\int_t^1 d t=\log |t|+C \\ & =\log |(x+\sin x)|+C \end{aligned}$$