Correct Answer - Option 1 :
\(\rm log\frac{ {|5+\sqrt {41}}|}{ {|4+\sqrt {32}}|}\)
Concept:
- \(\rm \int\frac{dx}{\sqrt {x^{2}+a^{2}}}=log{\left |x+\sqrt {x^{2}+a^{2}}\right |}+C\)
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\(\rm \int_{0}^{x}{f(x) dx}=F(x)-F(0)\), where F(x) is the anti-derivative of f(x).
Calculation:
Given: \(\rm \int_{4}^{5} \frac{dx}{\sqrt {x^{2}+16}}\)
Using the formula, \(\rm \int\frac{dx}{\sqrt {x^{2}+a^{2}}}=log{\left |x+\sqrt {x^{2}+a^{2}}\right |}+C\)
\(\Rightarrow \rm \int_{4}^{5} \frac{dx}{\sqrt {x^{2}+16}}=\left [ log{\left |x+\sqrt {x^{2}+16}\right |} \right ]_{4}^{5}\)
\(\Rightarrow \rm \int_{4}^{5} \frac{dx}{\sqrt {x^{2}+16}}=log{|5+\sqrt {5^{2}+16}}|-log{|4+\sqrt {4^{2}+16}}|\)
\(=log{|5+\sqrt {41}}|-log{|4+\sqrt {32}}|\)
It is known that log a - log b = log (a / b)
\(\Rightarrow \rm \int_{4}^{5} \frac{dx}{\sqrt {x^{2}-16}}=log\frac{ {|5+\sqrt {41}}|}{ {|4+\sqrt {32}}|}\)
Hence, the correct answer is option 1.