indefinite-integration

Question: $\int_{{}}^{{}}{\frac{1}{{{x}^{2}}\sqrt{1+{{x}^{2}}}}}\ dx=$

1) $-\frac{\sqrt{1+{{x}^{2}}}}{x}+c$

2) $\frac{\sqrt{1+{{x}^{2}}}}{x}+c$

3) $-\frac{\sqrt{1-{{x}^{2}}}}{x}+c$

4) $-\frac{\sqrt{{{x}^{2}}-1}}{x}+c$

Solution:

Explanation: No Explanation

Integration by substitution

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