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# Evaluate the following integrals:

Question:

Evaluate $\int \sqrt{3 \mathrm{x}^{2}+4 \mathrm{x}+1} \mathrm{dx}$

Solution:

Make perfect square of quadratic equation

$3 x^{2}+4 x+1=3\left(x^{2}+\frac{4}{3} x+\frac{1}{3}\right)$

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

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

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

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

Using formula, $\int \sqrt{x^{2}-a^{2}} d x=\frac{x}{2} \sqrt{x^{2}-a^{2}}-\frac{a^{2}}{2} \ln \left(x+\sqrt{x^{2}-a^{2}}\right)$

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

$y=\frac{3 x+2}{6} \sqrt{3 x^{2}+4 x+1}-\frac{\sqrt{3}}{18} \ln \left(\left(x+\frac{2}{3}\right)+\sqrt{x^{2}+\frac{4 x}{3}+\frac{1}{3}}\right)+c$