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Indefinite Integrals — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsIndefinite Integrals5 Marks
Question
Evaluate the following integrals:$\int\frac{2\text{x}}{2+\text{x}-\text{x}^2}\text{ dx}$
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Answer

$\int\frac{2\text{x}\text{ dx}}{(2+\text{x}-\text{x}^2)}$ $2\text{x}=\text{A}\frac{\text{d}}{\text{dx}}\big(2+\text{x}-\text{x}^2\big)+\text{B}$ $2\text{x}=\text{A}(0+1-2\text{x})+\text{B}$ $2\text{x}=(-2\text{A})\text{x}+\text{A}+\text{B}$ Comparing the coefficients of like power of x, $-2\text{A}=2$ $\text{A}=-1$ $\text{A}+\text{B}=0$ $-1+\text{B}=0$ $\text{B}=1$ Now, $\int\frac{2\text{x}\text{ dx}}{(2+\text{x}-\text{x}^2)}$ $=\int\Big(\frac{-1(1-2\text{x})+1}{-\text{x}^2+\text{x}+2}\Big)\text{dx}$ $=-\int\Big(\frac{1-2\text{x}}{-\text{x}^2+\text{x}+2}\Big)\text{dx}+\int\frac{\text{dx}}{-\text{x}^2+\text{x}+2}$ $=-\text{I}_1+\text{I}_2\ ....(1)$ (say) where $\text{I}_1=\int\Big(\frac{1-2\text{x}}{-\text{x}^2+\text{x}+2}\Big)\text{dx}$ $\text{I}_2=\int\frac{\text{dx}}{-\text{x}^2+\text{x}+2}$ $\text{I}_1=\int\Big(\frac{1-2\text{x}}{-\text{x}^2+\text{x}+2}\Big)\text{dx}$ Let $-\text{x}^2+\text{x}+2=\text{t}$ $\Rightarrow(1-2\text{x})\text{dx}=\text{dt}$ $\text{I}_1=\int\frac{\text{dt}}{\text{t}}$ $\text{I}_1=\log|\text{t}|+\text{C}_1\ ....(2)$ $\text{I}_2=\int\frac{\text{dx}}{-\text{x}^2+\text{x}+2}$ $\text{I}_2=\int\frac{-\text{dx}}{\text{x}^2+\text{x}+2}$ $\text{I}_2=\int\frac{-\text{dx}}{\text{x}^2-\text{x}+\big(\frac{1}{2}\big)^2-\big(\frac{1}{2}\big)^2-2}$ $\text{I}_2=\int\frac{-\text{dx}}{\big(\text{x}-\frac{1}{2}\big)^2-\big(\frac{3}{2}\big)^2}$ $\text{I}_2=-\frac{1}{2\times\frac{3}{2}}\log\Bigg|\frac{\text{x}-\frac{1}{2}-\frac{3}{2}}{\text{x}-\frac{1}{2}+\frac{3}{2}}\Bigg|+\text{C}_2$ $\text{I}_2=-\frac{1}{3}\log\Big|\frac{\text{x}-2}{\text{x}+1}\Big|+\text{C}_2\ ....(3)$ From (1) (2) and (3) $\int\Big(\frac{2\text{x}}{2+\text{x}+\text{x}^2}\Big)\text{ dx}=-\log\big|2+\text{x}-\text{x}^2\big|-\frac{1}{3}\log\Big|\frac{\text{x}-2}{\text{x}+1}\Big|+\text{C}_1+\text{C}_2$ $=-\log\big|2+\text{x}-\text{x}^2\big|+\frac{1}{3}\log\Big|\frac{1+\text{x}}{\text{x}-2}\Big|+\text{C}$Where, $\text{C}=\text{C}_1+\text{C}_2$

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