Evaluate the following integrals -
$\int(2 x+3) \sqrt{x^{2}+4 x+3} d x$
Let $I=\int(2 x+3) \sqrt{x^{2}+4 x+3} d x$
Let us assume $2 x+3=\lambda \frac{d}{d x}\left(x^{2}+4 x+3\right)+\mu$
$\Rightarrow 2 \mathrm{x}+3=\lambda\left[\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}^{2}\right)+\frac{\mathrm{d}}{\mathrm{dx}}(4 \mathrm{x})+\frac{\mathrm{d}}{\mathrm{dx}}(3)\right]+\mu$
$\Rightarrow 2 \mathrm{x}+3=\lambda\left[\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}^{2}\right)+4 \frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{x})+\frac{\mathrm{d}}{\mathrm{dx}}(3)\right]+\mu$
We know $\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}^{\mathrm{n}}\right)=\mathrm{nx}^{\mathrm{n}-1}$ and derivative of a constant is 0 .
$\Rightarrow 2 x+3=\lambda\left(2 x^{2-1}+4+0\right)+\mu$
$\Rightarrow 2 x+3=\lambda(2 x+4)+\mu$
$\Rightarrow 2 x+3=2 \lambda x+4 \lambda+\mu$
Comparing the coefficient of $x$ on both sides, we get
$2 \lambda=2 \Rightarrow \lambda=1$
Comparing the constant on both sides, we get
$4 \lambda+\mu=3$
$\Rightarrow 4(1)+\mu=3$
$\Rightarrow 4+\mu=3$
$\therefore \mu=-1$
Hence, we have $2 x+3=(2 x+4)-1$
Substituting this value in I, we can write the integral as
$I=\int[(2 x+4)-1] \sqrt{x^{2}+4 x+3} d x$
$\Rightarrow I=\int\left[(2 x+4) \sqrt{x^{2}+4 x+3}-\sqrt{x^{2}+4 x+3}\right] d x$
$\Rightarrow I=\int(2 x+4) \sqrt{x^{2}+4 x+3} d x-\int \sqrt{x^{2}+4 x+3} d x$
Let $I_{1}=\int(2 x+4) \sqrt{x^{2}+4 x+3} d x$
Now, put $x^{2}+4 x+3=t$
$\Rightarrow(2 x+4) d x=d t$ (Differentiating both sides)
Substituting this value in $\mathrm{I}_{1}$, we can write
$I_{1}=\int \sqrt{t} d t$
$\Rightarrow I_{1}=\int t^{\frac{1}{2}} d t$
Recall $\int x^{n} d x=\frac{x^{n+1}}{n+1}+c$
$\Rightarrow \mathrm{I}_{1}=\frac{\mathrm{t}^{\frac{1}{2}+1}}{\frac{1}{2}+1}+\mathrm{c}$
$\Rightarrow \mathrm{I}_{1}=\frac{\mathrm{t}^{\frac{3}{2}}}{\frac{3}{2}}+\mathrm{c}$
$\Rightarrow \mathrm{I}_{1}=\frac{2}{3} \mathrm{t}^{\frac{3}{2}}+\mathrm{c}$
$\therefore I_{1}=\frac{2}{3}\left(x^{2}+4 x+3\right)^{\frac{3}{2}}+c$
Let $I_{2}=-\int \sqrt{x^{2}+4 x+3} d x$
We can write $x^{2}+4 x+3=x^{2}+2(x)(2)+2^{2}-2^{2}+3$
$\Rightarrow x^{2}+4 x+3=(x+2)^{2}-4+3$
$\Rightarrow x^{2}+4 x+3=(x+2)^{2}-1$
$\Rightarrow x^{2}+4 x+3=(x+2)^{2}-1^{2}$
Hence, we can write $I_{2}$ as
$\mathrm{I}_{2}=-\int \sqrt{(\mathrm{x}+2)^{2}-1^{2}} \mathrm{dx}$
Recall $\int \sqrt{x^{2}-a^{2}} d x=\frac{x}{2} \sqrt{x^{2}-a^{2}}-\frac{a^{2}}{2} \ln \left|x+\sqrt{x^{2}-a^{2}}\right|+c$
$\Rightarrow \mathrm{I}_{2}=-\left[\frac{(\mathrm{x}+2)}{2} \sqrt{(\mathrm{x}+2)^{2}-1^{2}}-\frac{1^{2}}{2} \ln \left|(\mathrm{x}+2)+\sqrt{(\mathrm{x}+2)^{2}-1^{2}}\right|\right]+\mathrm{c}$
$\Rightarrow \mathrm{I}_{2}=-\left[\frac{(\mathrm{x}+2)}{2} \sqrt{\mathrm{x}^{2}+4 \mathrm{x}+3}-\frac{1}{2} \ln \left|\mathrm{x}+2+\sqrt{\mathrm{x}^{2}+4 \mathrm{x}+3}\right|\right]+\mathrm{c}$
$\therefore \mathrm{I}_{2}=-\frac{1}{2}(\mathrm{x}+2) \sqrt{\mathrm{x}^{2}+4 \mathrm{x}+3}+\frac{1}{2} \ln \left|\mathrm{x}+2+\sqrt{\mathrm{x}^{2}+4 \mathrm{x}+3}\right|+\mathrm{c}$
Substituting $I_{1}$ and $I_{2}$ in $I$, we get
$I=\frac{2}{3}\left(x^{2}+4 x+3\right)^{\frac{3}{2}}-\frac{1}{2}(x+2) \sqrt{x^{2}+4 x+3}+\frac{1}{2} \ln \left|x+2+\sqrt{x^{2}+4 x+3}\right|$
$+c$
Thus,$\int(2 x+3) \sqrt{x^{2}+4 x+3} d x=\frac{2}{3}\left(x^{2}+4 x+3\right)^{\frac{2}{2}}-\frac{1}{2}(x+2) \sqrt{x^{2}+4 x+3}+$
$\frac{1}{2} \ln \left|x+2+\sqrt{x^{2}+4 x+3}\right|+c$
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