Question

$\int (2x+3)\sqrt{x^2+3x+2}\mathrm{dx}$ is equal to.

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

Answer 1

The correct option is D

$\frac{2}{3}{\left(x^2+3x+2\right)}^{3/2}+C$
To solve these types of integrals we express the linear term as a sum of derivative of the quadratic term and a constant term.
Suppose the integral to be $I=\int (\mathrm{ax}+b)\sqrt{{\mathrm{cx}}^2+\mathrm{dx}+e}\mathrm{dx}$
Here, $(\mathrm{ax}+b)=A\frac{d}{\mathrm{dx}}({\mathrm{cx}}^2+\mathrm{dx}+e)+B$

For this case, $(2x+3)=A\frac{d}{\mathrm{dx}}(x^2+3x+2)+B$
$\Rightarrow (2x+3)=A(2x+3)+B$
Now, comparing the coefficients of $x$ and the constant term we get
$A=1$ and $B=0$.
Now we can write our integral as:
$I=\int (2x+3)\sqrt{x^2+3x+2}\mathrm{dx}$
$\Rightarrow I=\int \left(\frac{d}{\mathrm{dx}}(x^2+3x+2)\right\}\times \sqrt{x^2+3x+2}\mathrm{dx}$

Now we substitute $t=(x^2+3x+2)$
then, $\mathrm{dt}=(2x+3)\mathrm{dx}$
substituting these back in the integral we get,
$I=\int t^{1/2}\mathrm{dt}$
which gives $I=\frac{t^{3/2}}{3/2}=\frac{2}{3}{t}^{3/2}+C$
substituting back $t$ we get $I=\frac{2}{3}{\left(x^2+3x+2\right)}^{3/2}+C$