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# Evaluate the following integrals:

Question:

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

Solution:

Make perfect square of quadratic equation

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

$=3\left[\frac{4}{9}-\left(x^{2}-2\left(\frac{1}{3}\right)(x)+\left(\frac{1}{3}\right)^{2}\right)\right]$

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

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

Using formula, $\int \sqrt{a^{2}-x^{2}} d x=\frac{a^{2}}{2} \sin ^{-1} \frac{x}{a}+\frac{x}{2} \sqrt{a^{2}-x^{2}}$

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

$y=\frac{2 \sqrt{3}}{9} \sin ^{-1} \frac{(3 x-1)}{2}+\frac{(3 x-1)}{6} \sqrt{1+2 x-3 x^{2}}+c$