Question

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

• A. $\frac{2x-5}{4}\sqrt{1+5x-x^2}+\frac{13}{8}{\sin }^{-1}\left(\frac{2x-3}{\sqrt{13}}\right)+C$
• B. $\frac{2x-3}{4}\sqrt{1+3x-x^2}+\frac{13}{8}{\sin }^{-1}\left(\frac{2x-3}{\sqrt{13}}\right)+C$
• C. $\frac{3x-2}{4}\sqrt{1+3x-x^2}+\frac{13}{8}{\sin }^{-1}\left(\frac{2x-3}{\sqrt{13}}\right)+C$
• D. $\frac{2x-3}{4}\sqrt{1+2x-x^2}+\frac{13}{7}{\sin }^{-1}\left(\frac{2x-3}{\sqrt{13}}\right)+C$

Answer 1

The correct option is B $\frac{2x-3}{4}\sqrt{1+3x-x^2}+\frac{13}{8}{\sin }^{-1}\left(\frac{2x-3}{\sqrt{13}}\right)+C$

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

$=\int \sqrt{1+\frac{9}{4}-(x^2-2x\left(\frac{3}{2}\right)+\frac{9}{4})}dx$

$=\int \sqrt{{\left(\frac{\sqrt{13}}{2}\right)}^2-{(x-\frac{3}{2})}^2}dx$
Put $x-\frac{3}{2}=t$

$\Rightarrow dx=dt$
$I=\int \sqrt{{\left(\frac{\sqrt{13}}{2}\right)}^2-t^2}dt$
$=\frac{t}{2}\sqrt{\frac{13}{4}-t^2}+\frac{13}{4\times 2}{\sin }^{-1}\left(\frac{t}{\frac{\sqrt{13}}{2}}\right)+C$
$=\frac{x-\frac{3}{2}}{2}\sqrt{1+3x-x^2}+\frac{13}{4\times 2}{\sin }^{-1}\left(\frac{x-\frac{3}{2}}{\frac{\sqrt{13}}{2}}\right)+C$

$=\frac{2x-3}{4}\sqrt{1+3x-x^2}+\frac{13}{8}{\sin }^{-1}\left(\frac{2x-3}{\sqrt{13}}\right)+C$