# ∫ 2 X + 1 √ X 2 + 4 X + 3 D X - Mathematics

Sum
$\int\frac{2x + 1}{\sqrt{x^2 + 4x + 3}} \text{ dx }$

#### Solution

$\text{ Let I } = \int\frac{\left( 2x + 1 \right) dx}{\sqrt{x^2 + 4x + 3}}$
$\text{ Consider,}$
$2x + 1 = A \frac{d}{dx} \left( x^2 + 4x + 3 \right) + B$
$\Rightarrow 2x + 1 = A \left( 2x + 4 \right) + B$
$\Rightarrow 2x + 1 = \left( 2A \right) x + 4A + B$
$\text{Equating Coefficients of like terms}$
$\text{ 2 A} = 2$
$\Rightarrow A = 1$
$\text{ And }$
$4A + B = 1$
$\Rightarrow 4 + B = 1$
$\Rightarrow B = - 3$
$\therefore I = \int\left( \frac{2x + 4 - 3}{\sqrt{x^2 + 4x + 3}} \right)dx$
$= \int\frac{\left( 2x + 4 \right) dx}{\sqrt{x^2 + 4x + 3}} - 3\int\frac{dx}{\sqrt{x^2 + 4x + 4 - 4 + 3}}$
$= \int\frac{\left( 2x + 4 \right) dx}{\sqrt{x^2 + 4x + 3}} - 3\int\frac{dx}{\sqrt{\left( x + 2 \right)^2 - 1^2}}$
$\text{ Let x}^2 + 4x + 3 = t$
$\Rightarrow \left( 2x + 4 \right) dx = dt$
$\text{ Then,}$
$I = \int\frac{dt}{\sqrt{t}} - 3\int\frac{dx}{\sqrt{\left( x + 2 \right)^2 - 1^2}}$
$= \int t^{- \frac{1}{2}} dt - 3 \int\frac{dx}{\sqrt{\left( x + 2 \right)^2 - 1^2}}$
$= \left[ \frac{t^{- \frac{1}{2} + 1}}{- \frac{1}{2} + 1} \right] - 3 \text{ log }\left| x + 2 + \sqrt{\left( x + 2 \right)^2 - 1} \right| + C$
$= 2\sqrt{t} - 3 \text{ log} \left| x + 2 + \sqrt{x^2 + 4x + 3} \right| + C$
$= 2\sqrt{x^2 + 4x + 3} - 3 \text{ log} \left| x + 2 + \sqrt{x^2 + 4x + 3} \right| + C$

Concept: Indefinite Integral Problems
Is there an error in this question or solution?

#### APPEARS IN

RD Sharma Class 12 Maths
Chapter 19 Indefinite Integrals
Exercise 19.21 | Q 15 | Page 111