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Evaluate the following integrals -

Question:

Evaluate the following integrals -

$\int(2 x-4) \sqrt{x^{2}-4 x+3} d x$

Solution:

Let $I=\int(2 x-5) \sqrt{x^{2}-4 x+3} d x$

Let us assume $2 \mathrm{x}-5=\lambda \frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}^{2}-4 \mathrm{x}+3\right)+\mu$

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

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

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 .

$\Rightarrow 2 x-5=\lambda\left(2 x^{2-1}-4+0\right)+\mu$

$\Rightarrow 2 x-5=\lambda(2 x-4)+\mu$

$\Rightarrow 2 x-5=2 \lambda x+\mu-4 \lambda$

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

$2 \lambda=2 \Rightarrow \lambda=1$

Comparing the constant on both sides, we get

$\mu-4 \lambda=-5$

$\Rightarrow \mu-4(1)=-5$

$\Rightarrow \mu-4=-5$

$\therefore \mu=-1$

Hence, we have $2 x-5=(2 x-4)-1$

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

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

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

$\Rightarrow I=\int(2 x-4) \sqrt{x^{2}-4 x+3} d x-\int \sqrt{x^{2}-4 x+3} d x$

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

Now, put $x^{2}-4 x+3=t$

$\Rightarrow(2 x-4) d x=d t$ (Differentiating both sides)

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

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

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

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

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

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

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

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

Let $I_{2}=-\int \sqrt{x^{2}-4 x+3} d x$

We can write $x^{2}-4 x+3=x^{2}-2(x)(2)+2^{2}-2^{2}+3$

$\Rightarrow x^{2}-4 x+3=(x-2)^{2}-4+3$

$\Rightarrow x^{2}-4 x+3=(x-2)^{2}-1$

$\Rightarrow x^{2}-4 x+3=(x-2)^{2}-1^{2}$

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

$\mathrm{I}_{2}=-\int \sqrt{(\mathrm{x}-2)^{2}-1^{2}} \mathrm{dx}$

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

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

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

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

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

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

$+c$

Thus,$\int(2 x-5) \sqrt{x^{2}-4 x+3} d x=\frac{2}{3}\left(x^{2}-4 x+3\right)^{\frac{2}{2}}-\frac{1}{2}(x-2) \sqrt{x^{2}-4 x+3}+$

$\frac{1}{2} \ln \left|x-2+\sqrt{x^{2}-4 x+3}\right|+c$

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