# Evaluate the following integrals -

Question:

Evaluate the following integrals -

$\int(x+1) \sqrt{2 x^{2}+3} d x$

Solution:

Let $\mathrm{I}=\int(\mathrm{x}+1) \sqrt{2 \mathrm{x}^{2}+3} \mathrm{dx}$

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

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

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

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 .

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

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

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

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

$4 \lambda=1 \Rightarrow \lambda=\frac{1}{4}$

Comparing the constant on both sides, we get

$\mu=1$

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

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

$I=\int\left[\frac{1}{4}(4 x)+1\right] \sqrt{2 x^{2}+3} d x$

$\Rightarrow \mathrm{I}=\int\left[\frac{1}{2}(4 \mathrm{x}) \sqrt{2 \mathrm{x}^{2}+3}+\sqrt{2 \mathrm{x}^{2}+3}\right] \mathrm{dx}$

$\Rightarrow \mathrm{I}=\int \frac{1}{4}(4 \mathrm{x}) \sqrt{2 \mathrm{x}^{2}+3} \mathrm{~d} \mathrm{x}+\int \sqrt{2 \mathrm{x}^{2}+3} \mathrm{dx}$

$\Rightarrow \mathrm{I}=\frac{1}{4} \int(4 \mathrm{x}) \sqrt{2 \mathrm{x}^{2}+3} \mathrm{~d} \mathrm{x}+\int \sqrt{2 \mathrm{x}^{2}+3} \mathrm{dx}$

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

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

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

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

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

$\Rightarrow I_{1}=\frac{1}{4} \int t^{\frac{1}{2}} d t$

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

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

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

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

$\Rightarrow \mathrm{I}_{1}=\frac{1}{6} \mathrm{t}^{\frac{3}{2}}+\mathrm{c}$

$\therefore \mathrm{I}_{1}=\frac{1}{6}\left(2 \mathrm{x}^{2}+3\right)^{\frac{3}{2}}+\mathrm{c}$

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

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

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

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

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

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

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

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

$\Rightarrow I_{2}=\sqrt{2}\left[\frac{x}{2} \sqrt{x^{2}+\frac{3}{2}}+\frac{3}{4} \ln \left|x+\sqrt{x^{2}+\frac{3}{2}}\right|\right]+c$

$\Rightarrow I_{2}=\sqrt{2}\left[\frac{x}{2 \sqrt{2}} \sqrt{2 x^{2}+3}+\frac{3}{2 \times 2} \ln \left|x+\sqrt{x^{2}+\frac{3}{2}}\right|\right]+c$

$\therefore I_{2}=\frac{x}{2} \sqrt{2 x^{2}+3}+\frac{3}{2 \sqrt{2}} \ln \left|x+\sqrt{x^{2}+\frac{3}{2}}\right|+c$

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

$I=\frac{1}{6}\left(2 x^{2}+3\right)^{\frac{3}{2}}+\frac{x}{2} \sqrt{2 x^{2}+3}+\frac{3}{2 \sqrt{2}} \ln \left|x+\sqrt{x^{2}+\frac{3}{2}}\right|+$

Thus, $\int(\mathrm{x}+1) \sqrt{2 \mathrm{x}^{2}+3} \mathrm{dx}=\frac{1}{6}\left(2 \mathrm{x}^{2}+3\right)^{\frac{2}{2}}+\frac{\mathrm{x}}{2} \sqrt{2 \mathrm{x}^{2}+3}+\frac{3}{2 \sqrt{2}} \ln \left|\mathrm{x}+\sqrt{\mathrm{x}^{2}+\frac{3}{2}}\right|+\mathrm{c}$