# Evaluate the following integrals -

Question:

Evaluate the following integrals -

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

Solution:

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

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

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

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

$\Rightarrow 4 x+1=\lambda\left(2 x^{2-1}-1-0\right)+\mu$

$\Rightarrow 4 x+1=\lambda(2 x-1)+\mu$

$\Rightarrow 4 x+1=2 \lambda x+\mu-\lambda$

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

$2 \lambda=4 \Rightarrow \lambda=\frac{4}{2}=2$

Comparing the constant on both sides, we get

$\mu-\lambda=1$

$\Rightarrow \mu-2=1$

$\therefore \mu=3$

Hence, we have $4 x+1=2(2 x-1)+3$

Substituting this value in I, we can write the integral as

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

$\Rightarrow \mathrm{I}=\int\left[2(2 \mathrm{x}-1) \sqrt{\mathrm{x}^{2}-\mathrm{x}-2}+3 \sqrt{\mathrm{x}^{2}-\mathrm{x}-2}\right] \mathrm{dx}$

$\Rightarrow \mathrm{I}=\int 2(2 \mathrm{x}-1) \sqrt{\mathrm{x}^{2}-\mathrm{x}-2} \mathrm{~d} \mathrm{x}+\int 3 \sqrt{\mathrm{x}^{2}-\mathrm{x}-2} \mathrm{dx}$

$\Rightarrow \mathrm{I}=2 \int(2 \mathrm{x}-1) \sqrt{\mathrm{x}^{2}-\mathrm{x}-2} \mathrm{~d} \mathrm{x}+3 \int \sqrt{\mathrm{x}^{2}-\mathrm{x}-2} \mathrm{~d} \mathrm{x}$

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

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

$\Rightarrow(2 x-1) d x=d t$ (Differentiating both sides)

Substituting this value in $I_{1}$, we can write

$\mathrm{I}_{1}=2 \int \sqrt{\mathrm{t}} \mathrm{dt}$

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

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

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

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

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

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

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

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

We can write $x^{2}-x-2=x^{2}-2(x)\left(\frac{1}{2}\right)+\left(\frac{1}{2}\right)^{2}-\left(\frac{1}{2}\right)^{2}-2$

$\Rightarrow x^{2}-x-2=\left(x-\frac{1}{2}\right)^{2}-\frac{1}{4}-2$

$\Rightarrow x^{2}-x-2=\left(x-\frac{1}{2}\right)^{2}-\frac{9}{4}$

$\Rightarrow x^{2}-x-2=\left(x-\frac{1}{2}\right)^{2}-\left(\frac{3}{2}\right)^{2}$

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

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

$\left.\Rightarrow I_{2}=3\left[\frac{\left(x-\frac{1}{2}\right)}{2} \sqrt{\left(x-\frac{1}{2}\right)^{2}-\left(\frac{3}{2}\right)^{2}}-\frac{\left(\frac{3}{2}\right)^{2}}{2} \ln \mid\left(x-\frac{1}{2}\right)+\sqrt{\left(x-\frac{1}{2}\right)^{2}-\left(\frac{3}{2}\right)^{2}}\right]\right]$

$+c$

$\Rightarrow \mathrm{I}_{2}=3\left[\frac{2 \mathrm{x}-1}{4} \sqrt{\mathrm{x}^{2}-\mathrm{x}-2}-\frac{9}{8} \ln \left|\mathrm{x}-\frac{1}{2}+\sqrt{\mathrm{x}^{2}-\mathrm{x}-2}\right|\right]+\mathrm{c}$

$\therefore \mathrm{I}_{2}=\frac{3}{4}(2 \mathrm{x}-1) \sqrt{\mathrm{x}^{2}-\mathrm{x}-2}-\frac{27}{8} \ln \left|\mathrm{x}-\frac{1}{2}+\sqrt{\mathrm{x}^{2}-\mathrm{x}-2}\right|+\mathrm{c}$

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

$I=\frac{4}{3}\left(x^{2}-x-2\right)^{\frac{3}{2}}+\frac{3}{4}(2 x-1) \sqrt{x^{2}-x-2}-\frac{27}{8} \ln \left|x-\frac{1}{2}+\sqrt{x^{2}-x-2}\right|+c$

Thus, $\int(4 x+1) \sqrt{x^{2}-x-2} d x=\frac{4}{3}\left(x^{2}-x-2\right)^{\frac{2}{2}}+\frac{3}{4}(2 x-1) \sqrt{x^{2}-x-2}-$

$\frac{27}{8} \ln \left|x-\frac{1}{2}+\sqrt{x^{2}-x-2}\right|+c$