Question

$\int \left(\frac{t\sqrt{t}+\sqrt{t}}{t^2}\right)dt$ is equal to

• A. $2(\sqrt{t}-\frac{1}{\sqrt{t}})+c$
• B. $2(\sqrt{t}+\frac{1}{\sqrt{t}})+c$
• C. $2(\frac{1}{\sqrt{t}}-\sqrt{t})+c$
• D. $2(\sqrt{t}+2\sqrt{t})$

Answer 1

The correct option is A $2(\sqrt{t}-\frac{1}{\sqrt{t}})+c$

$\int \left(\frac{t\sqrt{t}+\sqrt{t}}{t^2}\right)dt=\int \frac{1}{\sqrt{t}}dt+\int \frac{\sqrt{t}}{t^2}\times \frac{\sqrt{t}}{\sqrt{t}}dt=\int \frac{1}{\sqrt{t}}dt+\int \frac{1}{t\sqrt{t}}dt$

$\int t^{-1/2}dt+\int t^{-3/2}dt=\frac{t^{(-\frac{1}{2}+1)}}{(-\frac{1}{2}+1)}+\frac{t^{(-\frac{3}{2}+1)}}{(-\frac{3}{2}+1)}$

$=2\sqrt{t}-\frac{2}{\sqrt{t}}+c=2(\sqrt{t}-\frac{1}{\sqrt{t}})+c$

$\int \left(\frac{t\sqrt{t}+\sqrt{t}}{t^2}\right)dt=2(\sqrt{t}-\frac{1}{\sqrt{t}})+c$