$\int \frac{2 x+5}{\sqrt{7-6 x-x^2}} \mathrm{~d} x=\mathrm{A} \sqrt{7-6 x-x^2}+\mathrm{B} \sin…

$\int \frac{2 x+5}{\sqrt{7-6 x-x^2}} \mathrm{~d} x=\mathrm{A} \sqrt{7-6 x-x^2}+\mathrm{B} \sin ^{-1}\left(\frac{x+3}{4}\right)+\mathrm{c}$ (where c c is a constant of integration) then the value of $A+B$ is
  1. -3
  2. 1
  3. -1
  4. 3

Solution

$\begin{aligned} & \text { Let } I=\int \frac{2 x+5}{\sqrt{7-6 x-x^2}} \mathrm{~d} x=\int \frac{2 x+6-6+5}{\sqrt{7-6 x-x^2}} \\ & =-1 \int \frac{-2 x-6}{\sqrt{7-6 x-x^2}} \mathrm{~d} x-\int \frac{1}{\sqrt{7+9-\left(9+6 x+x^2\right)}} \mathrm{d} x \\ & =-1 \int \frac{-2 x-6}{\sqrt{7-6 x-x^2}} \mathrm{~d} x-\int \frac{1}{\sqrt{(4)^2-(x+3)^2}} \mathrm{~d} x \end{aligned}$
Let $7-6 x-x^2=\mathrm{t}$ $\therefore \quad(-2 x-6) \mathrm{d} x=\mathrm{dt}$ $\begin{aligned} \therefore \quad \mathrm{I} & =-\int(\mathrm{t})^{\frac{-1}{2}} \mathrm{dt}-\sin ^{-1}\left(\frac{x+3}{4}\right)+\mathrm{c} \\ & =-2 \sqrt{7-6 x-x^2}-\sin ^{-1}\left(\frac{x+3}{4}\right)+\mathrm{c} \end{aligned}$ $\begin{array}{ll} \therefore & A=-2 \text { and } B=-1 \\ \therefore & A+B=-3 \end{array}$

Asked in: MHT CET 2024 (11 May Shift 2)

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