Question

$\int (2x+1)\sqrt{x^2+x+1}dx$ will be equal to -

• A. $\frac{2{(x^2+x+1)}^{3/2}}{3}+C$
  
  
• B. $\frac{2{(x^2+x+1)}^{3/2}}{7}+C$
• C. $\frac{2{(x^2+x+1)}^{3/2}}{9}+C$
• D. $\text{None of these}$

Answer 1

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



The given question is $\int (px+q)\sqrt{a{x}^2+bx+c}$ form.To approach these kind of questions we express px + q in terms of the derivative of the quadratic expression inside the under root.
So, px + q = $\alpha \frac{d}{dx}(a{x}^2+bx+c)+\beta$ Where $\alpha and\beta$ are the constants.
Here, we can see the given linear expression is already the derivative of the quadratic expression given
So, here $\alpha =1,\beta =0.$
$\int (2x+1)\sqrt{x^2+x+1}dx$
Let’s substitute $x^2+x+1=t^2$
So,$(2x+1)$dx = 2t dt
=$\int 2t.{\sqrt{t^2}}_{dt}$
Or$\int 2t.tdt$
= $\int 2t^{2}dt$
= $\frac{2t^3}{3}$ +C
Let’s put t =$\sqrt{x^2+x+1}$
So, the final answer would be $\frac{2{(x^2+x+1)}^{\frac{3}{2}}}{3}$ +C