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Evaluate the following integrals:

Question:

Evaluate the following integrals:

$\int \frac{2 x+1}{\sqrt{x^{2}+4 x+3}} d x$

Solution:

Given $I=\int \frac{2 x+1}{\sqrt{x^{2}+4 x+3}} d x$

Integral is of form $\int \frac{p x+q}{\sqrt{a x^{2}+b x+c}} d x$

Writing numerator as $\mathrm{px}+\mathrm{q}=\lambda\left\{\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{ax}^{2}+\mathrm{bx}+\mathrm{c}\right)\right\}+\mu$

$\Rightarrow p x+q=\lambda(2 a x+b)+\mu$

$\Rightarrow 2 x+1=\lambda(2 x+4)+\mu$

$\therefore \lambda=1$ and $\mu=-3$

Let $2 x+1=2 x+4-3$ and split,

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

$=2 \int \frac{x+2}{\sqrt{x^{2}+4 x+3}} d x-3 \int \frac{1}{\sqrt{x^{2}+4 x+3}} d x$

Consider $\int \frac{x+2}{\sqrt{x^{2}+4 x+3}} d x$

Let $u=x^{2}+4 x+3 \rightarrow d x=\frac{1}{2 x+4} d u$

$\Rightarrow \int \frac{\mathrm{x}+2}{\sqrt{\mathrm{x}^{2}+4 \mathrm{x}+3}} \mathrm{dx}=\int \frac{1}{2 \sqrt{\mathrm{u}}} \mathrm{du}$

$=\frac{1}{2} \int \frac{1}{\sqrt{\mathrm{u}}} \mathrm{du}$

We know that $\int \mathrm{x}^{\mathrm{n}} \mathrm{dx}=\frac{\mathrm{x}^{\mathrm{n}+1}}{\mathrm{n}+1}+\mathrm{c}$

$\Rightarrow \frac{1}{2} \int \frac{1}{\sqrt{u}} d u=\frac{1}{2}(2 \sqrt{u})$

$=\sqrt{u}=\sqrt{x^{2}+4 x+3}$

Consider $\int \frac{1}{\sqrt{x^{2}+4 x+3}} d x$

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

Let $u=x+2 \rightarrow d x=d u$

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

We know that $\int \frac{1}{\sqrt{x^{2}-1}} d x=\log \left(\sqrt{x^{2}-1}+x\right)+c$

$\Rightarrow \int \frac{1}{\sqrt{\mathrm{u}^{2}-1}} \mathrm{du}=\log \left(\sqrt{\mathrm{u}^{2}-1}+\mathrm{u}\right)$

$=\log \left(\sqrt{(\mathrm{x}+2)^{2}-1}+\mathrm{x}+2\right)$

Then,

$\Rightarrow \int \frac{2 x+1}{\sqrt{x^{2}+4 x+3}} d x=2 \int \frac{x+2}{\sqrt{x^{2}+4 x+3}} d x-3 \int \frac{1}{\sqrt{x^{2}+4 x+3}} d x$

$=2 \sqrt{x^{2}+4 x+3}-3 \log \left(\sqrt{(x+2)^{2}-1}+x+2\right)+c$

$=2 \sqrt{x^{2}+4 x+3}-3 \log \left(\sqrt{x^{2}+4 x+3}+x+2\right)+c$

$=2 \sqrt{(x+1)(x+3)}-3 \log (|\sqrt{(x+1)(x+3)}+x+2|)+c$

$\therefore \mathrm{I}=\int \frac{2 \mathrm{x}+1}{\sqrt{\mathrm{x}^{2}+4 \mathrm{x}+3}} \mathrm{dx}$

$=2 \sqrt{(\mathrm{x}+1)(\mathrm{x}+3)}-3 \log (|\sqrt{(\mathrm{x}+1)(\mathrm{x}+3)}+\mathrm{x}+2|)+\mathrm{c}$