int sqrt{1 + x^2} , dx = | vedclass

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Question

(A) We use the standard integration formula: $\int \sqrt{a^2 + x^2} \, dx = \frac{x}{2}\sqrt{a^2 + x^2} + \frac{a^2}{2}\ln|x + \sqrt{a^2 + x^2}| + C$.

Vedclass · Accepted Answer

$\frac{x}{2}\sqrt{1 + x^2} + \frac{1}{2}\ln|x + \sqrt{1 + x^2}| + C$

Vedclass · Answer

(A) We use the standard integration formula: $\int \sqrt{a^2 + x^2} \, dx = \frac{x}{2}\sqrt{a^2 + x^2} + \frac{a^2}{2}\ln|x + \sqrt{a^2 + x^2}| + C$. Substituting $a = 1$ into the formula: $\int \sqrt{1 + x^2} \, dx = \frac{x}{2}\sqrt{1 + x^2} + \frac{1^2}{2}\ln|x + \sqrt{1 + x^2}| + C$. Thus,the result is $\frac{x}{2}\sqrt{1 + x^2} + \frac{1}{2}\ln|x + \sqrt{1 + x^2}| + C$.

$\int \sqrt{1 + x^2} \, dx = $

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