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

(x + 1) \sqrt{x^{2} + 2 x + 5} + \frac{1}{2} l o g | x + 1 + \sqrt{x^{2} + 2 x + 5} | + C

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(x + 1) \sqrt{x^{2} + 2 x + 5} + 2 l o g | x + 1 + \sqrt{x^{2} + 2 x + 5} | + C

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(x + 1) \sqrt{x^{2} + 2 x + 5 - 2} l o g | x + 1 + \sqrt{x^{2} + 2 x + 5} | + C

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\frac{1}{2} (x + 1) \sqrt{x^{2} + 2 x + 5} + 2 l o g | x + 1 + \sqrt{x^{2} + 2 x + 5} | + C

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Solution

The correct option is D $\frac{1}{2} (x + 1) \sqrt{x^{2} + 2 x + 5} + 2 l o g | x + 1 + \sqrt{x^{2} + 2 x + 5} | + C$
Consider, $I = \int \sqrt{x^{2} + 2 x + 5} d x$
$I = \int \sqrt{x^{2} + 2 x + 1 + 4} d x$
$I = \int \sqrt{{(x + 1)}^{2} + 2^{2}} d x$
Now, $\int \sqrt{x^{2} + a^{2}} d x = \frac{x}{2} \sqrt{x^{2} + a^{2}} + \frac{a^{2}}{2} log ∣ ∣ x + \sqrt{x^{2} + a^{2}} ∣ ∣ + C$
$∴ I = \frac{(x + 1)}{2} \sqrt{{(x + 1)}^{2} + 2^{2}} + \frac{2^{2}}{2} log ∣ ∣ ∣ x + 1 + \sqrt{{(x + 1)}^{2} + 2^{2}} ∣ ∣ ∣ + C$
$I = \frac{1}{2} (x + 1) \sqrt{x^{2} + 2 x + 5} + 2 log ∣ ∣ x + 1 + \sqrt{x^{2} + 2 x + 5} ∣ ∣ + C$

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Similar questions

Evaluate $\int \frac{1}{(1 - x^{2}) \sqrt{1 + x^{2}}} d x$

\int \sqrt{(x^{2} + 2 x + 5)} d x

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

\int \frac{x + 3}{\sqrt{(x^{2} + 2 x + 2)}} d x = \sqrt{x^{2} + 2 x + 2} + 2 log [x + 1 - {(x + 1)^{2} + 1}^{1 / 2}] .

I = \int \frac{x^{3}}{\sqrt{x^{2} + 2 x + 5}} d x = \frac{1}{6} f (x) - \frac{A}{120} g (x) + C

f (x) = (2 x^{2} - 5 x - 36) \sqrt{x^{2} + 2 x + 5}, g (x) = log (x + 1 + \sqrt{x^{2} + 2 x + 5})

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