a.
Đặt [tex]\sqrt{x}=t\\\Rightarrow \frac{1}{2\sqrt{x}}dx=dt\\\Rightarrow dx=2t.dt[/tex]
Do đó:
[tex]\displaystyle \int \frac{(2\sqrt{x}-3)dx}{\sqrt{x}(1+\sqrt{x})}\\=\displaystyle \int \frac{(2t-3).2t.dt}{t(1+t)}\\=\displaystyle \int \frac{4t-6}{1+t}dt\\=\displaystyle \int (4-\frac{10}{t+1})dt\\=4t-10 \ln(t+1)+C\\=4\sqrt{x}-10 \ln(\sqrt{x}+1)+C[/tex]
b.
Đặt: [tex]\sqrt{x^2+4}=t\\\Rightarrow \frac{x}{\sqrt{x^2+4}}dx=dt\\\Rightarrow dx=\frac{\sqrt{x^2+4}}{x}dt[/tex]
Vậy:
[tex]\displaystyle \int \frac{dx}{x\sqrt{x^{2}+4}}\\=\displaystyle \int \frac{1}{t^2-4}dt\\=\displaystyle \int \frac{1}{4} .\frac{(t+2)-(t-2)}{(t+2)(t-2)}dt\\=\frac{1}{4} \displaystyle \int (\frac{1}{t-2}-\frac{1}{t+2})dt\\=\frac{1}{4}. \ln(t-2)-\frac{1}{4} \ln (t+2)+C\\=\frac{1}{4}. \ln(\sqrt{x^2+4}-2)-\frac{1}{4}.\ln (\sqrt{x^2+4}+2)+C[/tex]