Evaluate the following integrals -

Question:

Evaluate the following integrals -

$\int(x-3) \sqrt{x^{2}+3 x-18} d x$

Solution:

Let $I=\int(x-3) \sqrt{x^{2}+3 x-18} d x$

Let us assume $x-3=\lambda \frac{d}{d x}\left(x^{2}+3 x-18\right)+\mu$

$\Rightarrow x-3=\lambda\left[\frac{d}{d x}\left(x^{2}\right)+\frac{d}{d x}(3 x)-\frac{d}{d x}(18)\right]+\mu$

$\Rightarrow x-3=\lambda\left[\frac{d}{d x}\left(x^{2}\right)+3 \frac{d}{d x}(x)-\frac{d}{d x}(18)\right]+\mu$

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 .

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

$\Rightarrow x-3=\lambda(2 x+3)+\mu$

$\Rightarrow x-3=2 \lambda x+3 \lambda+\mu$

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

$3 \lambda+\mu=-3$

$\Rightarrow 3\left(\frac{1}{2}\right)+\mu=-3$

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

$\therefore \mu=-\frac{9}{2}$

Hence, we have $x-3=\frac{1}{2}(2 x+3)-\frac{9}{2}$

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

$I=\int\left[\frac{1}{2}(2 x+3)-\frac{9}{2}\right] \sqrt{x^{2}+3 x-18} d x$

$\Rightarrow \mathrm{I}=\int\left[\frac{1}{2}(2 \mathrm{x}+3) \sqrt{\mathrm{x}^{2}+3 \mathrm{x}-18}-\frac{9}{2} \sqrt{\mathrm{x}^{2}+3 \mathrm{x}-18}\right] \mathrm{dx}$

$\Rightarrow \mathrm{I}=\int \frac{1}{2}(2 \mathrm{x}+3) \sqrt{\mathrm{x}^{2}+3 \mathrm{x}-18} \mathrm{dx}-\int \frac{9}{2} \sqrt{\mathrm{x}^{2}+3 \mathrm{x}-18} \mathrm{dx}$

$\Rightarrow \mathrm{I}=\frac{1}{2} \int(2 \mathrm{x}+3) \sqrt{\mathrm{x}^{2}+3 \mathrm{x}-18} \mathrm{dx}-\frac{9}{2} \int \sqrt{\mathrm{x}^{2}+3 \mathrm{x}-18} \mathrm{dx}$

Let $I_{1}=\frac{1}{2} \int(2 x+3) \sqrt{x^{2}+3 x-18} d x$

Now, put $x^{2}+3 x-18=t$

$\Rightarrow(2 x+3) d x=d t$ (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 $\int x^{n} d x=\frac{x^{n+1}}{n+1}+c$

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

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

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

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

$\therefore \mathrm{I}_{1}=\frac{1}{3}\left(\mathrm{x}^{2}+3 \mathrm{x}-18\right)^{\frac{3}{2}}+\mathrm{c}$

Let $\mathrm{I}_{2}=-\frac{9}{2} \int \sqrt{\mathrm{x}^{2}+3 \mathrm{x}-18} \mathrm{dx}$

We can write $x^{2}+3 x-18=x^{2}+2(x)\left(\frac{3}{2}\right)+\left(\frac{3}{2}\right)^{2}-\left(\frac{3}{2}\right)^{2}-18$

$\Rightarrow x^{2}+3 x-18=\left(x+\frac{3}{2}\right)^{2}-\frac{9}{4}-18$

$\Rightarrow x^{2}+3 x-18=\left(x+\frac{3}{2}\right)^{2}-\frac{81}{4}$

$\Rightarrow x^{2}+3 x-18=\left(x+\frac{3}{2}\right)^{2}-\left(\frac{9}{2}\right)^{2}$

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

$\mathrm{I}_{2}=-\frac{9}{2} \int \sqrt{\left(\mathrm{x}+\frac{3}{2}\right)^{2}-\left(\frac{9}{2}\right)^{2}} \mathrm{dx}$

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

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

$\left.-\frac{\left(\frac{9}{2}\right)^{2}}{2} \ln \left|\left(x+\frac{3}{2}\right)+\sqrt{\left(x+\frac{3}{2}\right)^{2}-\left(\frac{9}{2}\right)^{2}}\right|\right]+c$

$\Rightarrow \mathrm{I}_{2}=-\frac{9}{2}\left[\frac{(2 \mathrm{x}+3)}{4} \sqrt{\mathrm{x}^{2}+3 \mathrm{x}-18}-\frac{81}{8} \ln \left|\mathrm{x}+\frac{3}{2}+\sqrt{\mathrm{x}^{2}+3 \mathrm{x}-18}\right|\right]+\mathrm{c}$

$\therefore \mathrm{I}_{2}=-\frac{9}{8}(2 \mathrm{x}+3) \sqrt{\mathrm{x}^{2}+3 \mathrm{x}-18}+\frac{729}{16} \ln \left|\mathrm{x}+\frac{3}{2}+\sqrt{\mathrm{x}^{2}+3 \mathrm{x}-18}\right|+\mathrm{c}$

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

$I=\frac{1}{3}\left(x^{2}+3 x-18\right)^{\frac{3}{2}}-\frac{9}{8}(2 x+3) \sqrt{x^{2}+3 x-18}$

$+\frac{729}{16} \ln \left|x+\frac{3}{2}+\sqrt{x^{2}+3 x-18}\right|+c$

Thus,$\int(x-3) \sqrt{x^{2}+3 x-18} d x=\frac{1}{3}\left(x^{2}+3 x-18\right)^{\frac{3}{2}}-\frac{9}{8}(2 x+3) \sqrt{x^{2}+3 x-18}+$

$\frac{729}{16} \ln \left|x+\frac{3}{2}+\sqrt{x^{2}+3 x-18}\right|+c$

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