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NCERT Class XII Mathematics Chapter - - Solutions

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Exercise - 7.10
Evaluate the definite integrals in Exercises 198 to 205 using substitution.
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Question : 203 of 271
Marks: +1, -0
∫02dxx+4−x2\int\limits_{0}^{2} \frac{dx}{x+4-x^2}
Solution:  
Let I=∫02dxx+4−x2I=\int\limits_{0}^{2} \frac{dx}{x+4-x^2} =∫02dx4−(x2−x)=\int\limits_{0}^{2} \frac{dx}{4-(x^2-x)}
=∫02dx4+14−(x2−x+14)=\int\limits_{0}^{2} \frac{dx}{4+\frac{1}{4}-\left(x^2-x+\frac{1}{4}\right)}
=∫02dx174−(x−12)2=\int\limits_{0}^{2} \frac{dx}{\frac{17}{4}-\left(x-\frac{1}{2}\right)^2}
=∫02dx(172)2−(x−12)2=\int\limits_{0}^{2} \frac{dx}{\left(\frac{\sqrt{17}}{2}\right)^2-\left(x-\frac{1}{2}\right)^2}
=12⋅172log⁡∣172+(x−12)172−(x−12)∣02=\frac{1}{2 \cdot \frac{\sqrt{17}}{2}} \log \left| \frac{\frac{\sqrt{17}}{2}+\left(x-\frac{1}{2}\right)}{\frac{\sqrt{17}}{2}-\left(x-\frac{1}{2}\right)} \right|_0^2
=117log⁡∣17+2x−117−2x+1∣02=\frac{1}{\sqrt{17}} \log \left| \frac{\sqrt{17}+2x-1}{\sqrt{17}-2x+1} \right|_0^2
=117[log⁡(17+4−117−4+1)−log⁡(17−117+1)]=\frac{1}{\sqrt{17}} \left[ \log \left( \frac{\sqrt{17}+4-1}{\sqrt{17}-4+1} \right) - \log \left( \frac{\sqrt{17}-1}{\sqrt{17}+1} \right) \right]
=117[log⁡(17+4−117−4+1)−log⁡(17−117+1)]=\frac{1}{\sqrt{17}} \left[ \log \left( \frac{\sqrt{17}+4-1}{\sqrt{17}-4+1} \right) - \log \left( \frac{\sqrt{17}-1}{\sqrt{17}+1} \right) \right]
=117log⁡[17+317−3⋅17+117−1]=\frac{1}{\sqrt{17}} \log \left[ \frac{\sqrt{17}+3}{\sqrt{17}-3} \cdot \frac{\sqrt{17}+1}{\sqrt{17}-1} \right]
=117log⁡(20+41720−417)=\frac{1}{\sqrt{17}} \log \left( \frac{20+4\sqrt{17}}{20-4\sqrt{17}} \right)
=117log⁡(21+5174)=\frac{1}{\sqrt{17}} \log \left( \frac{21+5\sqrt{17}}{4} \right)
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