Question

$\int_{}^{}{(2x + 3)\sqrt{x^{2} + 4x + 3}}dx =$

• A. $\log|(x + 2) + (\sqrt{x^{2} + 4x + 3)}| + c$
• B. $\log|(x + 2) + (\sqrt{x^{2} + 4x + 3)}| + c$
• C. $\log|(x - 2) + (\sqrt{x^{2} + 4x + 3)}| + c$
• D. None of these

Answer 1

Answer: (D)

None of these

Answer 2

Let $2x + 3 = M\frac{d}{dx}(x^{2} + 4x + 3) + N$

$\Rightarrow$ $2x + 3 = M(2x + 4) + N$

Equating the coefficients of x and constant terms on both sides, we get

$2 = 2M \Rightarrow M = 1$ and $3 = 4M + N \Rightarrow N = 3 - 4 \times 1 = - 1\therefore 2x + 3 = (2x + 4) - 1$

Hence,$I = \int_{}^{}{\lbrack(2x + 4) - 1\rbrack}\sqrt{x^{2} + 4x + 3dx}$ $= \int_{}^{}{(2x + 4)\sqrt{x^{2} + 4x + 3}}dx - \int_{}^{}\sqrt{x^{2} + 4x + 3}dx$ $= I_{1} - I_{2},(say)$

Now, $I_{1} = \int_{}^{}{(2x + 4)\sqrt{x^{2} + 4x + 3}dx}$

Putting $x^{2} + 4x + 3 = t \Rightarrow (2x + 4)dx = dt,$ we have

$I_{1} = \int_{}^{}t^{1/2}dt = \frac{t^{3/2}}{3/2} + c_{1}$ $= \frac{2}{3}(x^{2} + 4x + 3)^{3/2} + c_{1}$

$I_{2} = \int_{}^{}\sqrt{x^{2} + 4x + 3}dx = \int_{}^{}\sqrt{(x + 2)^{2} - 1^{2}}dx$

$= \frac{1}{2}(x + 2)\sqrt{(x + 2)^{2} - 1^{2}} - \frac{1}{2}.1^{2}\log\left| x + 2 + \sqrt{(x + 2)^{2} - 1^{2}} \right| + c_{2} = \frac{1}{2}(x + 2)\sqrt{x^{2} + 4x + 3} - \frac{1}{2}\log\left| x + 2 + \sqrt{x^{2} + 4x + 3} \right| + c_{2}$

$\therefore I = I_{1} - I_{2}$

$= \frac{2}{3}(x^{2} + 4x + 3)^{3/2} - \left\lbrack \frac{1}{2}(x + 2)\sqrt{x^{2} + 4x + 3} \right.\ - \frac{1}{2}\log\left. \ \left| x + 2 + \sqrt{x^{2} + 4x + 3} \right| \right\rbrack + c,$ (where, $c = c_{1} - c_{2}),$