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Miscellaneous Exercise 4 · Q79

Q.Evaluate: ∫04[x2+2x+3]−1dx\int_0^4 \left[\sqrt{x^2+2x+3}\right]^{-1}dx

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∫04dx(x+1)2+2=[ln⁡∣(x+1)+(x+1)2+2∣]04.\int_0^4\frac{dx}{\sqrt{(x+1)^2+2}}=\Big[\ln\big|(x+1)+\sqrt{(x+1)^2+2}\big|\Big]_0^4.

At x=4x=4: 5+27=5+335+\sqrt{27}=5+3\sqrt3. At x=0x=0: 1+31+\sqrt3.

ln⁡(5+33)−ln⁡(1+3)=ln⁡5+331+3.\ln(5+3\sqrt3)-\ln(1+\sqrt3)=\ln\frac{5+3\sqrt3}{1+\sqrt3}. …

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