Question

If $x=(6y+4)(3y^2+4y+3)$, then $\int xdy$ will be

• A. $\frac{1}{3y^2+4y+3}+C$
• B. $\frac{(3y^2+4y+3{)}^2}{2}+C$
• C. $3y^2+4y+3+C$
• D. $\frac{(6y+4)}{3y^2+4y+3}+C$

Answer 1

The correct option is B $\frac{(3y^2+4y+3{)}^2}{2}+C$

Given $x=(6y+4)(3y^2+4y+3)$
Let, $t=3y^2+4y+3$
Differentiate w.r.t. to y,
$\frac{dt}{dy}=(6y+4)$
$\Rightarrow (6y+4)dy=dt$
Then $\int xdy=\int tdt=\frac{t^2}{2}+C=\frac{(3y^2+4y+3{)}^2}{2}+C$