Question

Integrate the function: $\sqrt{1+3x-x^2}$

Answer 1

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

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

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

=$\int\sqrt{\bigg(\frac{\sqrt13}{2}\bigg)^2-\bigg(x-\frac{3}{2}\bigg)^2}dx$

It is known that,$\int\sqrt{a^2-x^2}dx=\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}\sin^{-1}\frac{x}{a}+C$

∴$I= \frac{x-\frac{3}{2}}{2}\sqrt{1+3x-x^2}+\frac{13}{4*2}\sin^{-1}\bigg(\frac{x-\frac{3}{2}}{\frac{\sqrt 13}{2}}\bigg)+C$

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