Question

Evaluate $\int \sqrt{1+y^2}\cdot 2ydy$.

• A. $I=\frac{2}{3}(1+y^2{)}^{3/2}+C$
  
• B. $I=\frac{2}{5}(1-y^2{)}^{3/2}+C$
• C. $I=\frac{2}{3}(1-y^2{)}^{3/2}+C$
• D. None of these.
  

Answer 1

The correct option is A $I=\frac{2}{3}(1+y^2{)}^{3/2}+C$

Let $I=\int \sqrt{1+y^2}\cdot 2ydy$
Let $u=1+y^2$, then $du=2ydy$.
$I=\int u^{1/2}du$
$=\frac{u^{1/2}+1}{\left(1/2\right)+1}$ [Integrate, using rule no. $3$ with $n=1/2$]
$=\frac{2}{3}{u}^{3/2}+C$ [Simpler form]
$=\frac{2}{3}(1+y^2{)}^{3/2}+C$ [Replace $u$ by $1+y^2]$.