Question

If $I=\int (7x+4)\sqrt{5-4x-2x^2}dx,$ then I equals

• A. $\frac{7}{2}x\sqrt{5-4x-2x^2}+\frac{21\sqrt{2}}{4}{\sin }^{-1}\left(\frac{\sqrt{2}(x+1)}{\sqrt{7}}\right)+C$
• B. $(7x+3)\sqrt{5-4x-2x^2}+\frac{21\sqrt{2}}{4}{\sin }^{-1}\left(\frac{\sqrt{2}(x+1)}{\sqrt{7}}\right)+C$
• C. $\frac{1}{6}(14x^2-3x-4)\sqrt{5-4x-2x^2}+\frac{21\sqrt{2}}{4}{\sin }^{-1}\left(\sqrt{2}(x+1)\right)+C$
• D. none of these

Answer 1

The correct option is D none of these

Let $I=\int (7x+4)\sqrt{5-4x-2x^2}dx$
Put $t=x+\frac{b}{2a}=x+\frac{-4}{-2}=x+1$
$\Rightarrow dx=dt,7x+4=7(t-1)+4=7t+3,5-4x-2x^2=5-4(t-1)-2{(t-1)}^2=7-2t^2$
$I=\int (7t+3)\sqrt{7-2t^2}dt=7\int t\sqrt{7-2t^2}dt+3\int \sqrt{7-2t^2}dt$
$=7\left(\frac{2}{3}\right)(-\frac{1}{4}){(7-2t^2)}^{3/2}$
$+3\sqrt{2}(\frac{t}{2}\sqrt{\frac{7}{2}-t^2}+\frac{1}{2}\left(\frac{7}{2}\right){\sin }^{-1}\left(\frac{t}{\sqrt{7/2}}\right))+c$
$=-\frac{7}{6}(7-2t^2)+\frac{3t}{2}\sqrt{7-2t^2}+\frac{21\sqrt{2}}{4}{\sin }^{-1}\left(\frac{\sqrt{2}t}{\sqrt{7}}\right)+c$
$=-\frac{7}{6}{(5-4x-2x^2)}^{3/2}+\frac{3}{2}(x-1){(5-4x-2x^2)}^{1/2}+\frac{21\sqrt{2}}{4}{\sin }^{-1}\left(\frac{\sqrt{2}(x+1)}{\sqrt{7}}\right)+c$