Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSIntegrals1 Mark
Question
Find the integral: $\int \frac{x+3}{\sqrt{5-4 x-x^{2}}} d\ x$
✓
Answer
Given integral is:$\int \frac{x+3}{\sqrt{5-4 x-x^{2}}} d x$
Let us express
$x+3=A \frac{d}{d x}\left(5-4 x-x^{2}\right)+B = A (- 4 - 2x) + B$
Equating the coefficients of $x$ and the constant terms from both sides, we get
$- 2A = 1$ and $-4 A + B = 3,$ i.e., $A=-\frac{1}{2}$ and $B = 1$
Therefore, $\int \frac{x+3}{\sqrt{5-4 x-x^{2}}} d x$
$=-\frac{1}{2} \int \frac{(-4-2 x) d x}{\sqrt{5-4 x-x^{2}}}+\int \frac{d x}{\sqrt{5-4 x-x^{2}}}$
= $-\frac{1}{2} I_{1}+I_{2} ........(i)$
In $I_1,$ put $5 - 4x - x^2 = t,$
so that $(- 4 - 2x) dx = dt.$
Therefore, $I_{1}=\int \frac{(-4-2 x) d x}{\sqrt{5-4 x-x^{2}}}$
$=\int \frac{d t}{\sqrt{t}}$
$=2 \sqrt{t}+C_{1}$
$=2 \sqrt{5-4 x-x^{2}}+\mathrm{C}_{1} .......(ii)$
Now consider $I_{2}=\int \frac{d x}{\sqrt{5-4 x-x^{2}}}$
$=\int \frac{d x}{\sqrt{9-(x+2)^{2}}}$
Put $x + 2 = t,$
so that $dx = dt.$
Therefore, $I_{2}=\int \frac{d t}{\sqrt{3^{2}-t^{2}}}$
$=\sin ^{-1} \frac{t}{3}+C_{2}$
$=\sin ^{-1} \frac{x+2}{3}+C_{2} .......(iii)$
Substituting $(ii)$ and $(iii)$ in $(i),$ we obtain
$\int \frac{x+3}{\sqrt{5-4 x-x^{2}}}$
$=-\sqrt{5-4 x-x^{2}}+\sin ^{-1} \frac{x+2}{3}+\mathrm{C}, $
where $\mathrm{C}=\mathrm{C}_{2}-\frac{\mathrm{C}_{1}}{2}$
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