Question

Solve: $\int \frac{dx}{\sqrt{x+3}-\sqrt{x+2}}$

• A. $\frac{2}{3}\left[\right(x+3{)}^{\frac{3}{2}}+(x+2{)}^{\frac{3}{2}}]+C$
• B. $-\frac{2}{3}\left[\right(x+3{)}^{\frac{3}{2}}+(x+2{)}^{\frac{3}{2}}]+C$
• C. $-\frac{2}{3}\left[\right(x+3{)}^{\frac{3}{2}}-(x+2{)}^{\frac{3}{2}}]+C$
  
• D. $\frac{2}{3}\left[\right(x+3{)}^{\frac{3}{2}}-(x+2{)}^{\frac{3}{2}}]+C$
  

Answer 1

The correct option is A $\frac{2}{3}\left[\right(x+3{)}^{\frac{3}{2}}+(x+2{)}^{\frac{3}{2}}]+C$

Let $I=\int \frac{dx}{\sqrt{(x+3)}-\sqrt{(x+2)}}$

$=\int \frac{\sqrt{(x+3)}+\sqrt{(x+2)}}{x+3-x-2}dx$

$=\int \frac{\sqrt{(x+3)}+\sqrt{(x+2)}}{1}dx$

$=\int \sqrt{(x+3)}dx+\int \sqrt{(x+2)}dx$

Put $x+3=t,dx=dt$

$x+2=u,dx=du$

$\Rightarrow I=\int (t{)}^{\frac{1}{2}}dt+\int (u{)}^{\frac{1}{2}}du$

$\Rightarrow I=\frac{2}{3}(t{)}^{\frac{3}{2}}+\frac{2}{3}(u{)}^{\frac{3}{2}}+C$

$\Rightarrow I=\frac{2}{3}\left[\right(x+3{)}^{\frac{3}{2}}+(x+2{)}^{\frac{3}{2}}]+c$