Question

$\int \frac{x+2}{(x^2+4x+3)}dx$ is?

• A. $\ln (x^2+4x+3)+C$
  
  
• B. $\frac{1}{2}\ln (x^2+4x+3)+C$
• C. $\frac{1}{4}\ln (x^2+4x+3)+C$
  
  
• D. None of these

Answer 1

The correct option is B

$\frac{1}{2}\ln (x^2+4x+3)+C$

Consider the given integral.

$I=\int \frac{x+2}{(x^2+4x+3)}dx$

Let

$t=x^2+4x+3$

$\frac{dt}{dx}=2x+4+0$

$\frac{dt}{dx}=2(x+2)$

$\frac{dt}{2}=(x+2)dx$

Therefore,

$I=\frac{1}{2}\int \frac{1}{t}dt$

$I=\frac{1}{2}\ln \left(t\right)+C$

On putting the value of $t$, we get

$I=\frac{1}{2}\ln (x^2+4x+3)+C$

Hence, this is the answer.