The correct option is C (x+2)√(x^2+4x+5)+ln |x+2+√(x^2+4x+5)|2+C Let I=∫√(x^2+4x+5)dx =∫√(x^2+4x+4+1)dx = ∫√((x+2)^2+1)dx ∵∫√(x^2+a^2) ...

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Functions

Solve : ∫√x...

Question

Solve : $\int \sqrt{x^{2} + 4 x + 5} d x$

\frac{(x + 2)^{2} \sqrt{x^{2} + 4 x + 5} + ln | x + 2 + \sqrt{x^{2} + 4 x + 5} |}{2} + C

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\frac{(x + 2) \sqrt{x^{2} + 4 x + 5} - ln | x + 2 + \sqrt{x^{2} + 4 x + 5} |}{2} + C

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\frac{(x + 2) \sqrt{x^{2} + 4 x + 5} + ln | x + 2 + \sqrt{x^{2} + 4 x + 5} |}{2} + C

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Solution

The correct option is C $\frac{(x + 2) \sqrt{x^{2} + 4 x + 5} + ln | x + 2 + \sqrt{x^{2} + 4 x + 5} |}{2} + C$
Let $I = \int \sqrt{x^{2} + 4 x + 5} d x$
$= \int \sqrt{x^{2} + 4 x + 4 + 1} d x$
$= \int \sqrt{(x + 2)^{2} + 1} d x$
$∵ \int \sqrt{x^{2} + a^{2}} d x = \frac{x}{2} \sqrt{x^{2} + a^{2}} + \frac{a^{2}}{2} log | x + \sqrt{x^{2} + a^{2}} | + C$
In $I$ , $a = 1$ and $x = x + 2$
$∴ I = \frac{(x + 2) \sqrt{x^{2} + 4 x + 5} + ln | x + 2 + \sqrt{x^{2} + 4 x + 5} | + C}{2}$

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x

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