int sqrt{1+x^{2}} , dx is equal to | vedclass

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(D) We use the standard integration formula for $\int \sqrt{x^{2}+a^{2}} \, dx$: $\int \sqrt{x^{2}+a^{2}} \, dx = \frac{x}{2} \sqrt{x^{2}+a^{2}} + \fr

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) We use the standard integration formula for $\int \sqrt{x^{2}+a^{2}} \, dx$: $\int \sqrt{x^{2}+a^{2}} \, dx = \frac{x}{2} \sqrt{x^{2}+a^{2}} + \frac{a^{2}}{2} \log |x+\sqrt{x^{2}+a^{2}}| + C$ Comparing $\int \sqrt{1+x^{2}} \, dx$ with the standard form,we have $a = 1$. Substituting $a = 1$ into the formula: $\int \sqrt{1+x^{2}} \, dx = \frac{x}{2} \sqrt{1+x^{2}} + \frac{1^{2}}{2} \log |x+\sqrt{1+x^{2}}| + C$ $= \frac{x}{2} \sqrt{1+x^{2}} + \frac{1}{2} \log |x+\sqrt{1+x^{2}}| + C$ Thus,the correct option is $D$.

$\int \sqrt{1+x^{2}} \, dx$ is equal to

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