Evaluate the following integrals -

Let $I=\int x \sqrt{x^{2}+x} d x$

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

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

We know $\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}^{\mathrm{n}}\right)=\mathrm{nx}^{\mathrm{n}-1}$

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

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

$\Rightarrow x=2 \lambda x+\lambda+\mu$

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

$2 \lambda=1 \Rightarrow \lambda=\frac{1}{2}$

Comparing the constant on both sides, we get

$\lambda+\mu=0$

$\Rightarrow \frac{1}{2}+\mu=0$

$\therefore \mu=-\frac{1}{2}$

Hence, we have $x=\frac{1}{2}(2 x+1)-\frac{1}{2}$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$\Rightarrow I_{1}=\frac{1}{2} \times \frac{2}{3} t^{\frac{3}{2}}+c$

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

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

Let $I_{2}=-\frac{1}{2} \int \sqrt{x^{2}+x} d x$

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

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

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

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

Recall $\int \sqrt{\mathrm{x}^{2}-\mathrm{a}^{2}} \mathrm{dx}=\frac{\mathrm{x}}{2} \sqrt{\mathrm{x}^{2}-\mathrm{a}^{2}}-\frac{\mathrm{a}^{2}}{2} \ln \left|\mathrm{x}+\sqrt{\mathrm{x}^{2}-\mathrm{a}^{2}}\right|+\mathrm{c}$

$\Rightarrow \mathrm{I}_{2}=-\frac{1}{2}\left[\frac{\left(\mathrm{x}+\frac{1}{2}\right)}{2} \sqrt{\left(\mathrm{x}+\frac{1}{2}\right)^{2}-\left(\frac{1}{2}\right)^{2}}\right.$

$\left.-\frac{\left(\frac{1}{2}\right)^{2}}{2} \ln \left|\left(x+\frac{1}{2}\right)+\sqrt{\left(x+\frac{1}{2}\right)^{2}-\left(\frac{1}{2}\right)^{2}}\right|\right]+c$

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

$\therefore \mathrm{I}_{2}=-\frac{1}{8}(2 \mathrm{x}+1) \sqrt{\mathrm{x}^{2}+\mathrm{x}}+\frac{1}{16} \ln \left|\mathrm{x}+\frac{1}{2}+\sqrt{\mathrm{x}^{2}+\mathrm{x}}\right|+\mathrm{c}$

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

$I=\frac{1}{3}\left(x^{2}+x\right)^{\frac{3}{2}}-\frac{1}{8}(2 x+1) \sqrt{x^{2}+x}+\frac{1}{16} \ln \left|x+\frac{1}{2}+\sqrt{x^{2}+x}\right|+c$

Thus, $\int \mathrm{x} \sqrt{\mathrm{x}^{2}+\mathrm{x}} \mathrm{dx}=\frac{1}{3}\left(\mathrm{x}^{2}+\mathrm{x}\right)^{\frac{3}{2}}-\frac{1}{8}(2 \mathrm{x}+1) \sqrt{\mathrm{x}^{2}+\mathrm{x}}+\frac{1}{16} \ln \left|\mathrm{x}+\frac{1}{2}+\sqrt{\mathrm{x}^{2}+\mathrm{x}}\right|+\mathrm{c}$