# Evaluate the following integrals -

Question:

Evaluate the following integrals -

$\int(x-2) \sqrt{2 x^{2}-6 x+5} d x$

Solution:

Let $I=\int(x-2) \sqrt{2 x^{2}-6 x+5} d x$

Let us assume $x-2=\lambda \frac{d}{d x}\left(2 x^{2}-6 x+5\right)+\mu$

$\Rightarrow \mathrm{x}-2=\lambda\left[\frac{\mathrm{d}}{\mathrm{dx}}\left(2 \mathrm{x}^{2}\right)-\frac{\mathrm{d}}{\mathrm{dx}}(6 \mathrm{x})-\frac{\mathrm{d}}{\mathrm{dx}}(5)\right]+\mu$

$\Rightarrow \mathrm{x}-2=\lambda\left[2 \frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}^{2}\right)-6 \frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{x})-\frac{\mathrm{d}}{\mathrm{dx}}(5)\right]+\mu$

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

$\Rightarrow x-2=\lambda\left(2 \times 2 x^{2-1}-6-0\right)+\mu$

$\Rightarrow x-2=\lambda(4 x-6)+\mu$

$\Rightarrow x-2=4 \lambda x+\mu-6 \lambda$

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

$\mu-6 \lambda=-2$

$\Rightarrow \mu-6\left(\frac{1}{4}\right)=-2$

$\Rightarrow \mu-\frac{3}{2}=-2$

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

Hence, we have $x-2=\frac{1}{4}(4 x-6)-\frac{1}{2}$

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

$\mathrm{I}=\int\left[\frac{1}{4}(4 \mathrm{x}-6)-\frac{1}{2}\right] \sqrt{2 \mathrm{x}^{2}-6 \mathrm{x}+5} \mathrm{dx}$

$\Rightarrow \mathrm{I}=\int\left[\frac{1}{4}(4 \mathrm{x}-6) \sqrt{2 \mathrm{x}^{2}-6 \mathrm{x}+5}-\frac{1}{2} \sqrt{2 \mathrm{x}^{2}-6 \mathrm{x}+5}\right] \mathrm{dx}$

$\Rightarrow \mathrm{I}=\int \frac{1}{4}(4 \mathrm{x}-6) \sqrt{2 \mathrm{x}^{2}-6 \mathrm{x}+5} \mathrm{dx}-\int \frac{1}{2} \sqrt{2 \mathrm{x}^{2}-6 \mathrm{x}+5} \mathrm{dx}$

$\Rightarrow \mathrm{I}=\frac{1}{4} \int(4 \mathrm{x}-6) \sqrt{2 \mathrm{x}^{2}-6 \mathrm{x}+5} \mathrm{dx}-\frac{1}{2} \int \sqrt{2 \mathrm{x}^{2}-6 \mathrm{x}+5} \mathrm{dx}$

Let $\mathrm{I}_{1}=\frac{1}{4} \int(4 \mathrm{x}-6) \sqrt{2 \mathrm{x}^{2}-6 \mathrm{x}+5} \mathrm{dx}$

Now, put $2 x^{2}-6 x+5=t$

$\Rightarrow(4 x-6) d x=d t$ (Differentiating both sides)

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

$I_{1}=\frac{1}{4} \int \sqrt{t} d t$

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

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

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

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

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

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

$\therefore \mathrm{I}_{1}=\frac{1}{6}\left(2 \mathrm{x}^{2}-6 \mathrm{x}+5\right)^{\frac{3}{2}}+\mathrm{c}$

Let $I_{2}=-\frac{1}{2} \int \sqrt{2 x^{2}-6 x+5} d x$

We can write $2 x^{2}-6 x+5=2\left(x^{2}-3 x+\frac{5}{2}\right)$

$\Rightarrow 2 x^{2}-6 x+5=2\left[x^{2}-2(x)\left(\frac{3}{2}\right)+\left(\frac{3}{2}\right)^{2}-\left(\frac{3}{2}\right)^{2}+\frac{5}{2}\right]$

$\Rightarrow 2 x^{2}-6 x+5=2\left[\left(x-\frac{3}{2}\right)^{2}-\frac{9}{4}+\frac{5}{2}\right]$

$\Rightarrow 2 x^{2}-6 x+5=2\left[\left(x-\frac{3}{2}\right)^{2}+\frac{1}{4}\right]$

$\Rightarrow 2 x^{2}-6 x+5=2\left[\left(x-\frac{3}{2}\right)^{2}+\left(\frac{1}{2}\right)^{2}\right]$

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

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

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

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

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

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

$\left.+\frac{\left(\frac{1}{2}\right)^{2}}{2} \ln \left|\left(x-\frac{3}{2}\right)+\sqrt{\left(x-\frac{3}{2}\right)^{2}+\left(\frac{1}{2}\right)^{2}}\right|\right]+c$

$\Rightarrow I_{2}=-\frac{1}{\sqrt{2}}\left[\frac{2 x-3}{4} \sqrt{x^{2}-3 x+\frac{5}{2}}+\frac{1}{8} \ln \left|x-\frac{3}{2}+\sqrt{x^{2}-3 x+\frac{5}{2}}\right|\right]+c$

$\Rightarrow I_{2}=-\frac{1}{\sqrt{2}}\left[\frac{2 x-3}{4 \sqrt{2}} \sqrt{2 x^{2}-6 x+5}+\frac{1}{8} \ln \left|x-\frac{3}{2}+\sqrt{x^{2}-3 x+\frac{5}{2}}\right|\right]+c$

$\therefore I_{2}=-\frac{1}{8}(2 x-3) \sqrt{2 x^{2}-6 x+5}-\frac{1}{8 \sqrt{2}} \ln \left|x-\frac{3}{2}+\sqrt{x^{2}-3 x+\frac{5}{2}}\right|+c$

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

$I=\frac{1}{6}\left(2 x^{2}-6 x+5\right)^{\frac{3}{2}}-\frac{1}{8}(2 x-3) \sqrt{2 x^{2}-6 x+5}$

$-\frac{1}{8 \sqrt{2}} \ln \left|x-\frac{3}{2}+\sqrt{x^{2}-3 x+\frac{5}{2}}\right|+c$

Thus, $\int(x-2) \sqrt{2 x^{2}-6 x+5} d x=\frac{1}{6}\left(2 x^{2}-6 x+5\right)^{\frac{2}{2}}-\frac{1}{8}(2 x-3) \sqrt{2 x^{2}-6 x+5}-$

$\frac{1}{8 \sqrt{2}} \ln \left|x-\frac{3}{2}+\sqrt{x^{2}-3 x+\frac{5}{2}}\right|+c$