Question
Evaluate : $\int \frac{x+2}{\sqrt{x^2+2 x+3}}$
✓
Answer
Let $I =\int \frac{x+2}{\sqrt{x^2+2 x+3}}$
$x +2=A \frac{d}{d z}\left[ x ^2+2 x +3\right]+ B$
$\Rightarrow x +2=2 Ax +2 A+ B$
Comparing the coefficients, we have $, 2A = 1$ and $2A + B = 2$
$\Rightarrow A =\frac{1}{2}$
Substituting the value of $A$ in $2 A+B=2$, we have, $2 \times \frac{1}{2}+B=2$
$\Rightarrow 1+B=2$
$\Rightarrow B=2-1$
$\Rightarrow B=1$
Thus we have, $x +2=\frac{1}{2}[2 x+2]+1$
Hence, using values of $A,$ and $B,$ we have
$I=\int \frac{x+2}{\sqrt{x^2+2 x+3}} d x$
$\left.=\int \frac{\left[\frac{1}{2}[2 x+2]+1\right]}{\sqrt{x^2+2 x+3}}\right] d x$
$=\int \frac{\left[\frac{1}{2}[2 x+2]\right]}{\sqrt{x^2+2 x+3}} d x+\int \frac{d x}{\sqrt{x^2+2 x+3}}$
$=\frac{1}{2} \int \frac{[2 x+2]}{\sqrt{x^2+2 x+3}} d x+\int \frac{d x}{\sqrt{x^2+2 x+3}}$
Substituting $t = x ^2+2 x +3$ and $dt =2 x +2$
in the first integrand, we have, $I =\frac{1}{2} \int \frac{d t}{\sqrt{t}}+\int \frac{d x}{\sqrt{x^2+2 x+3}}$
$=\frac{1}{2} \times 2 \sqrt{t}+\int \frac{d x}{\sqrt{x^2+2 x+1+2}}+C$
$=\sqrt{t}+\int \frac{d x}{\sqrt{(x+1)^2+(\sqrt{2})^2}}+c$
$ I =\sqrt{x^2+2 x+3}+\log \left[| x +1|+\sqrt{(x+1)^2+(\sqrt{2})^2}\right]+C$
$\Rightarrow I =\sqrt{x^2+2 x+3}+\log \left[| x +1| \sqrt{x^2+2 x+3}\right]+ c $
$x +2=A \frac{d}{d z}\left[ x ^2+2 x +3\right]+ B$
$\Rightarrow x +2=2 Ax +2 A+ B$
Comparing the coefficients, we have $, 2A = 1$ and $2A + B = 2$
$\Rightarrow A =\frac{1}{2}$
Substituting the value of $A$ in $2 A+B=2$, we have, $2 \times \frac{1}{2}+B=2$
$\Rightarrow 1+B=2$
$\Rightarrow B=2-1$
$\Rightarrow B=1$
Thus we have, $x +2=\frac{1}{2}[2 x+2]+1$
Hence, using values of $A,$ and $B,$ we have
$I=\int \frac{x+2}{\sqrt{x^2+2 x+3}} d x$
$\left.=\int \frac{\left[\frac{1}{2}[2 x+2]+1\right]}{\sqrt{x^2+2 x+3}}\right] d x$
$=\int \frac{\left[\frac{1}{2}[2 x+2]\right]}{\sqrt{x^2+2 x+3}} d x+\int \frac{d x}{\sqrt{x^2+2 x+3}}$
$=\frac{1}{2} \int \frac{[2 x+2]}{\sqrt{x^2+2 x+3}} d x+\int \frac{d x}{\sqrt{x^2+2 x+3}}$
Substituting $t = x ^2+2 x +3$ and $dt =2 x +2$
in the first integrand, we have, $I =\frac{1}{2} \int \frac{d t}{\sqrt{t}}+\int \frac{d x}{\sqrt{x^2+2 x+3}}$
$=\frac{1}{2} \times 2 \sqrt{t}+\int \frac{d x}{\sqrt{x^2+2 x+1+2}}+C$
$=\sqrt{t}+\int \frac{d x}{\sqrt{(x+1)^2+(\sqrt{2})^2}}+c$
$ I =\sqrt{x^2+2 x+3}+\log \left[| x +1|+\sqrt{(x+1)^2+(\sqrt{2})^2}\right]+C$
$\Rightarrow I =\sqrt{x^2+2 x+3}+\log \left[| x +1| \sqrt{x^2+2 x+3}\right]+ c $
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