Question

Find $a,b$ in $\int \frac{x+2}{(x^2+3x+3)\sqrt{x+1}}dx=\frac{a}{\sqrt{b}}{\tan }^{-1}\left\{\frac{x}{\sqrt{3(x+1)}}\right\}+C$.

• A. $a=2$
• B. $b=3$
• C. $a=3$
• D. $b=2$

Answer 1

Answer: (A), (B)

$b=3$
$a=2$

Given:
$\int \frac{(x+2)dx}{(x^2+3x+3)\sqrt{x+1}}=\frac{9}{\sqrt{6}}{\tan }^{-1}\left|\frac{x}{\sqrt{3(x+1)}}\right|+C$

$Tofind=a\&bsolution;t^2=x+12tdt=(x+1){(x+1)}^2=t^4=x^2+2x+1=t^4{x}^2+3x+3=x^2+2x+1+x+2={(x+1)}^2+(x+2)\therefore I=\int \frac{t^2+1}{\{t^4+(t^2+1\left)\right\}.t}.2tdt=2\int \frac{t^2+1}{t^4+t^2+1}dt=2\int \frac{1+\frac{1}{t^2}}{t^2+1+\frac{1}{t^2}}dt\left[dividingnumereter\&denominaterby{t}^2\right]$
$=2\int \frac{(t+\frac{1}{t^2})dt}{{(t-\frac{1}{t})}^2+3}Takey=t-\frac{1}{t}dy=(t+\frac{1}{t^2})dtI=2\int \frac{dy}{y^2+3}=\frac{2}{\sqrt{3}}{\tan }^{-1}\left(\frac{y}{\sqrt{3}}\right)+CI=\frac{2}{\sqrt{3}}{\tan }^{-1}\left(\frac{t-\frac{1}{t}}{\sqrt{3}}\right)+C=\frac{2}{\sqrt{3}}{\tan }^{-1}\left(\frac{x}{\sqrt{3(x+1)}}\right)+C$
Hence the correct answer is
a$=$2 b$=$3