int sqrt{x^{2}+2 x+5} , dx is equal to | vedclass

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(A) We need to evaluate the integral $ I = \int \sqrt{x^{2}+2 x+5} \, dx $. First,complete the square for the quadratic expression: $ x^{2}+2x+5 = (x+

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$ \frac{1}{2}(x+1) \sqrt{x^{2}+2 x+5}+2 \log \left|x+1+\sqrt{x^{2}+2 x+5}\right|+C $

Vedclass · Answer

(A) We need to evaluate the integral $ I = \int \sqrt{x^{2}+2 x+5} \, dx $. First,complete the square for the quadratic expression: $ x^{2}+2x+5 = (x+1)^{2} + 4 = (x+1)^{2} + 2^{2} $. Now the integral becomes $ I = \int \sqrt{(x+1)^{2} + 2^{2}} \, dx $. Using the standard formula $ \int \sqrt{t^{2}+a^{2}} \, dt = \frac{t}{2} \sqrt{t^{2}+a^{2}} + \frac{a^{2}}{2} \log \left|t + \sqrt{t^{2}+a^{2}}\right| + C $,where $ t = x+1 $ and $ a = 2 $. Substituting these values: $ I = \frac{x+1}{2} \sqrt{(x+1)^{2} + 2^{2}} + \frac{2^{2}}{2} \log \left|(x+1) + \sqrt{(x+1)^{2} + 2^{2}}\right| + C $. Simplifying the expression: $ I = \frac{1}{2}(x+1) \sqrt{x^{2}+2x+5} + 2 \log \left|x+1 + \sqrt{x^{2}+2x+5}\right| + C $. Thus,the correct option is $ A $.

$ \int \sqrt{x^{2}+2 x+5} \, dx $ is equal to

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