Evaluate the following integrals -

Let $I=\int(3 x+1) \sqrt{4-3 x-2 x^{2}} d x$

Let us assume $3 x+1=\lambda \frac{d}{d x}\left(4-3 x-2 x^{2}\right)+\mu$

$\Rightarrow 3 x+1=\lambda\left[\frac{d}{d x}(4)-\frac{d}{d x}(3 x)-\frac{d}{d x}\left(2 x^{2}\right)\right]+\mu$

$\Rightarrow 3 x+1=\lambda\left[\frac{d}{d x}(4)-3 \frac{d}{d x}(x)-2 \frac{d}{d x}\left(x^{2}\right)\right]+\mu$

We know $\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}^{\mathrm{n}}\right)=\mathrm{n} \mathrm{x}^{\mathrm{n}-1}$ and derivative of a constant is 0 .

$\Rightarrow 3 x+1=\lambda\left(0-3-2 \times 2 x^{2-1}\right)+\mu$

$\Rightarrow 3 x+1=\lambda(-3-4 x)+\mu$

$\Rightarrow 3 x+1=-4 \lambda x+\mu-3 \lambda$

Comparing the coefficient of $x$ on both sides, we get

$-4 \lambda=3 \Rightarrow \lambda=-\frac{3}{4}$

Comparing the constant on both sides, we get

$\mu-3 \lambda=1$

$\Rightarrow \mu-3\left(-\frac{3}{4}\right)=1$

$\Rightarrow \mu+\frac{9}{4}=1$

$\therefore \mu=-\frac{5}{4}$

Hence, we have $3 x+1=-\frac{3}{4}(-3-4 x)-\frac{5}{4}$

Substituting this value in $\mathrm{I}$, we can write the integral as

$I=\int\left[-\frac{3}{4}(-3-4 x)-\frac{5}{4}\right] \sqrt{4-3 x-2 x^{2}} d x$

$\Rightarrow \mathrm{I}=\int\left[-\frac{3}{4}(-3-4 \mathrm{x}) \sqrt{4-3 \mathrm{x}-2 \mathrm{x}^{2}}-\frac{5}{4} \sqrt{4-3 \mathrm{x}-2 \mathrm{x}^{2}}\right] \mathrm{dx}$

$\Rightarrow \mathrm{I}=-\int \frac{3}{4}(-3-4 \mathrm{x}) \sqrt{4-3 \mathrm{x}-2 \mathrm{x}^{2}} \mathrm{dx}-\int \frac{5}{4} \sqrt{4-3 \mathrm{x}-2 \mathrm{x}^{2}} \mathrm{dx}$

$\Rightarrow \mathrm{I}=-\frac{3}{4} \int(-3-4 \mathrm{x}) \sqrt{4-3 \mathrm{x}-2 \mathrm{x}^{2}} \mathrm{dx}-\frac{5}{4} \int \sqrt{4-3 \mathrm{x}-2 \mathrm{x}^{2}} \mathrm{dx}$

Let $I_{1}=-\frac{3}{4} \int(-3-4 x) \sqrt{4-3 x-2 x^{2}} d x$

Now, put $4-3 x-2 x^{2}=t$

$\Rightarrow(-3-4 x) d x=d t$ (Differentiating both sides)

Substituting this value in $\mathrm{I}_{1}$, we can write

$\mathrm{I}_{1}=-\frac{3}{4} \int \sqrt{\mathrm{t} \mathrm{dt}}$

$\Rightarrow \mathrm{I}_{1}=-\frac{3}{4} \int \mathrm{t}^{\frac{1}{2}} \mathrm{dt}$

$\mathrm{I}_{1}=-\frac{3}{4} \int \sqrt{\mathrm{t}} \mathrm{dt}$

$\Rightarrow \mathrm{I}_{1}=-\frac{3}{4} \int \mathrm{t}^{\frac{1}{2}} \mathrm{dt}$

Recall $\int x^{n} d x=\frac{x^{n+1}}{n+1}+c$

$\Rightarrow \mathrm{I}_{1}=-\frac{3}{4}\left(\frac{\mathrm{t}^{\frac{1}{2}+1}}{\frac{1}{2}+1}\right)+\mathrm{c}$

$\Rightarrow \mathrm{I}_{1}=-\frac{3}{4}\left(\frac{\mathrm{t}^{\frac{3}{2}}}{\frac{3}{2}}\right)+\mathrm{c}$

$\Rightarrow \mathrm{I}_{1}=-\frac{3}{4} \times \frac{2}{3} \mathrm{t} \mathrm{t}^{\frac{3}{2}}+\mathrm{c}$

$\Rightarrow \mathrm{I}_{1}=-\frac{1}{2} \mathrm{t}^{\frac{3}{2}}+\mathrm{c}$

$\therefore \mathrm{I}_{1}=-\frac{1}{2}\left(4-3 \mathrm{x}-2 \mathrm{x}^{2}\right)^{\frac{3}{2}}+\mathrm{c}$

Let $I_{2}=-\frac{5}{4} \int \sqrt{4-3 x-2 x^{2}} d x$

We can write $4-3 x-2 x^{2}=-\left(2 x^{2}+3 x-4\right)$

$\Rightarrow 4-3 x-2 x^{2}=-2\left[x^{2}+\frac{3}{2} x-2\right]$

$\Rightarrow 4-3 x-2 x^{2}=-2\left[x^{2}+2(x)\left(\frac{3}{4}\right)+\left(\frac{3}{4}\right)^{2}-\left(\frac{3}{4}\right)^{2}-2\right]$

$\Rightarrow 4-3 x-2 x^{2}=-2\left[\left(x+\frac{3}{4}\right)^{2}-\frac{9}{16}-2\right]$

$\Rightarrow 4-3 x-2 x^{2}=-2\left[\left(x+\frac{3}{4}\right)^{2}-\frac{41}{16}\right]$

$\Rightarrow 4-3 x-2 x^{2}=2\left[\frac{41}{16}-\left(x+\frac{3}{4}\right)^{2}\right]$

$\Rightarrow 4-3 x-2 x^{2}=2\left[\left(\frac{\sqrt{41}}{4}\right)^{2}-\left(x+\frac{3}{4}\right)^{2}\right]$

Hence, we can write $I_{2}$ as

$I_{2}=-\frac{5}{4} \int \sqrt{2\left[\left(\frac{\sqrt{41}}{4}\right)^{2}-\left(x+\frac{3}{4}\right)^{2}\right]} d x$

$\Rightarrow I_{2}=-\frac{5 \sqrt{2}}{4} \int \sqrt{\left(\frac{\sqrt{41}}{4}\right)^{2}-\left(x+\frac{3}{4}\right)^{2}} d x$

Recall $\int \sqrt{a^{2}-x^{2}} d x=\frac{x}{2} \sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2} \sin ^{-1} \frac{x}{a}+c$

$\left.\Rightarrow I_{2}=-\frac{5 \sqrt{2}}{4}\left[\frac{\left(x+\frac{3}{4}\right)}{2}\right]{\left(\frac{\sqrt{41}}{4}\right)^{2}-\left(x+\frac{3}{4}\right)^{2}}+\frac{\left(\frac{\sqrt{41}}{4}\right)^{2}}{2} \sin ^{-1}\left(\frac{x+\frac{3}{4}}{\frac{\sqrt{41}}{4}}\right)\right]+c$

$\left.\Rightarrow I_{2}=-\frac{5 \sqrt{2}}{4}\left[\frac{(4 x+3)}{8}\right] \sqrt{2-\frac{3}{2} x-x^{2}}+\frac{41}{32} \sin ^{-1}\left(\frac{4 x+3}{\sqrt{41}}\right)\right]+c$

$\Rightarrow \mathrm{I}_{2}=-\frac{5 \sqrt{2}}{32}(4 \mathrm{x}+3) \sqrt{2-\frac{3}{2} \mathrm{x}-\mathrm{x}^{2}}-\frac{205 \sqrt{2}}{128} \sin ^{-1}\left(\frac{4 \mathrm{x}+3}{\sqrt{41}}\right)+\mathrm{c}$

$\therefore \mathrm{I}_{2}=-\frac{5}{32}(4 \mathrm{x}+3) \sqrt{4-3 \mathrm{x}-2 \mathrm{x}^{2}}-\frac{205 \sqrt{2}}{128} \sin ^{-1}\left(\frac{4 \mathrm{x}+3}{\sqrt{41}}\right)+\mathrm{c}$

Substituting $I_{1}$ and $I_{2}$ in $I$, we get

$\begin{aligned} I=-\frac{1}{2}(4-&\left.3 x-2 x^{2}\right)^{\frac{3}{2}}-\frac{5}{32}(4 x+3) \sqrt{4-3 x-2 x^{2}}-\frac{205 \sqrt{2}}{128} \sin ^{-1}\left(\frac{4 x+3}{\sqrt{41}}\right) \\ &+c \end{aligned}$

Thus,$\int(3 x+1) \sqrt{4-3 x-2 x^{2}} d x=-\frac{1}{2}\left(4-3 x-2 x^{2}\right)^{\frac{2}{2}}-\frac{5}{32}(4 x+$ 3) $\sqrt{4-3 x-2 x^{2}}-\frac{205 \sqrt{2}}{128} \sin ^{-1}\left(\frac{4 x+3}{\sqrt{41}}\right)+c$