$\int \sqrt{x^2+2x+5}\,dx$ equals
A$\dfrac{1}{2}(x+1)\sqrt{x^2+2x+5}+2\log\left|x+1+\sqrt{x^2+2x+5}\right|+C$
B$\dfrac{1}{2}(x+1)\sqrt{x^2+2x+5}+2\sin^{-1}\!\left(\dfrac{x+1}{2}\right)+C$
C$\dfrac{1}{2}(x+1)\sqrt{x^2+2x+5}-2\log\left|x+1+\sqrt{x^2+2x+5}\right|+C$
D$(x+1)\sqrt{x^2+2x+5}+C$
Answer & Solution
Correct answer: A. $\dfrac{1}{2}(x+1)\sqrt{x^2+2x+5}+2\log\left|x+1+\sqrt{x^2+2x+5}\right|+C$
1. Complete the square: $x^2+2x+5=(x+1)^2+4$.
2. Put $y=x+1$: integral $=\int \sqrt{y^2+2^2}\,dy$.
3. Use $\int \sqrt{y^2+a^2}\,dy=\dfrac{y}{2}\sqrt{y^2+a^2}+\dfrac{a^2}{2}\log|y+\sqrt{y^2+a^2}|$ with $a=2$.
4. Back-substitute: gives option A. Option B wrongly uses $\sin^{-1}$ (that is the $a^2-x^2$ form).
_Source: NCERT Class 12 Mathematics Ch 7 "Integrals", p.331_
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