Question

The value of the integral ${\int }_1^2\left(\frac{1}{x^2+4x+4}\right)dx$ is

• A. 1/12
• B. 4
• C. 1
• D. 6

Answer 1

The correct option is A 1/12

Let $I=\underset{1}{\overset{2}{\int }}\frac{dx}{x^2+4x+4}$

Now, $\int \frac{dx}{x^2+4x+4}=\int \frac{dx}{(x+2{)}^2}$

Let $x+2=t$

$dx=dt$

Therefore,

$\int \frac{dx}{x^2+4x+4}=\int \frac{dt}{t^2}$.............$[\because x^2+4x+4=(x+2{)}^2=t^2]$

$=\int t^{-2}dt$

$=\frac{t^{-1}}{-1}+C=-\frac{1}{(x+2)}$

Therefore,

$I=\underset{1}{\overset{2}{\int }}\frac{-1}{x+2}$

$I=-\frac{1}{2+2}+\frac{1}{1+2}$

$I=\frac{1}{3}-\frac{1}{4}=\frac{1}{12}$