Question

Evaluate the following integrals:

$\int \left(x+3\right)\sqrt{3-4x-x^2}\mathrm{d}x$

Answer 1

$$\begin{gathered} \mathrm{Let}I=\int \left(x+3\right)\sqrt{3-4x-x^2}\mathrm{d}x \\ \mathrm{We}\mathrm{express}x+3=A\left(\frac{d}{dx}\left(3-4x-x^2\right)\right)+B \\ x+3=A(-4-2x)+B \\ \mathrm{Equating}\mathrm{the}\mathrm{coefficients}\mathrm{of}x\mathrm{and}\mathrm{constants},\mathrm{we}\mathrm{get} \\ 1=-2A\mathrm{and}3=-4A+B \\ \mathrm{or}A=-\frac{1}{2}\mathrm{and}B=1 \\ \therefore I=\int \left(-\frac{1}{2}\left(-4-2x\right)+1\right)\sqrt{3-4x-x^2}\mathrm{d}x \\ =-\frac{1}{2}\int \left(-4-2x\right)\sqrt{3-4x-x^2}\mathrm{d}x+\int \sqrt{3-4x-x^2}\mathrm{d}x \\ =-\frac{1}{2}{I}_1+I_2...\left(1\right) \\ \mathrm{Now},I_1=\int \left(-4-2x\right)\sqrt{3-4x-x^2}\mathrm{d}x \\ \mathrm{Let}3-4x-x^2=u \\ \mathrm{On}\mathrm{differentiating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ \left(-4-2x\right)\mathrm{d}x=\mathrm{d}u \\ \therefore I_1=\int \sqrt{u}\mathrm{d}u \\ =\frac{2}{3}{u}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+c_1 \\ =\frac{2}{3}{\left(3-4x-x^2\right)}^{\frac{3}{2}}+c_1...\left(2\right) \\ \mathrm{And},I_2=\int \sqrt{3-4x-x^2}\mathrm{d}x \\ =\int \sqrt{3+4-4-4x-x^2}\mathrm{d}x \\ =\int \sqrt{{\left(\sqrt{7}\right)}^2-{\left(x+2\right)}^2}\mathrm{d}x \\ \mathrm{Let}\left(x+2\right)=u \\ \mathrm{On}\mathrm{differentiating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ \mathrm{d}x=\mathrm{d}u \\ \therefore I_2=\int \sqrt{{\left(\sqrt{7}\right)}^2-{\left(u\right)}^2}\mathrm{d}u \\ =\frac{u}{2}\sqrt{{\left(\sqrt{7}\right)}^2-{\left(u\right)}^2}+\frac{1}{2}{\left(\sqrt{7}\right)}^2{\sin }^{-1}\left(\frac{u}{\sqrt{7}}\right)+c_2 \\ =\frac{x+{\displaystyle 2}}{2}\sqrt{7-{\left(x+2\right)}^2}+\frac{7}{2}{\sin }^{-1}\left(\frac{x+2}{\sqrt{7}}\right)+c_2 \\ =\frac{x+{\displaystyle 2}}{2}\sqrt{3-4x-x^2}+\frac{7}{2}{\sin }^{-1}\left(\frac{x+2}{\sqrt{7}}\right)+c_2...\left(3\right) \\ \mathrm{From}\left(1\right),\left(2\right)\mathrm{and}\left(3\right),\mathrm{we}\mathrm{get} \\ \therefore I=-\frac{1}{2}\left(\frac{2}{3}{\left(3-4x-x^2\right)}^{\frac{3}{2}}+c_1\right)+\left(\frac{x+{\displaystyle 2}}{2}\sqrt{3-4x-x^2}+\frac{7}{2}{\sin }^{-1}\left(\frac{x+2}{\sqrt{7}}\right)+c_2\right) \\ =-\frac{1}{3}{\left(3-4x-x^2\right)}^{\frac{3}{2}}+\frac{x+{\displaystyle 2}}{2}\sqrt{3-4x-x^2}+\frac{7}{2}{\sin }^{-1}\left(\frac{x+2}{\sqrt{7}}\right)+c \\ \mathrm{Hence},\int \left(x+3\right)\sqrt{3-4x-x^2}\mathrm{d}x=-\frac{1}{3}{\left(3-4x-x^2\right)}^{\frac{3}{2}}+\frac{x+{\displaystyle 2}}{2}\sqrt{3-4x-x^2}+\frac{7}{2}{\sin }^{-1}\left(\frac{x+2}{\sqrt{7}}\right)+c \end{gathered}$$