Gujarat 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 I1, put 5 - 4x - x2 = 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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