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Find the value of \(\rm \int_{4}^{5} \frac{dx}{\sqrt {x^{2}+16}}\)
1. \(\rm log\frac{ {|5+\sqrt {41}}|}{ {|4+\sqrt {32}}|}\)
2. \(\rm log\frac{ {|5-\sqrt {41}}|}{ {|4+\sqrt {32}}|}\)
3. \(\rm log\frac{ {|5+\sqrt {41}}|}{ {|4-\sqrt {32}}|}\)
4. \(\rm log\frac{ {|5-\sqrt {41}}|}{ {|4-\sqrt {32}}|}\)
5. None of these

1 Answer

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Correct Answer - Option 1 : \(\rm log\frac{ {|5+\sqrt {41}}|}{ {|4+\sqrt {32}}|}\)

Concept:

  • \(\rm \int\frac{dx}{\sqrt {x^{2}+a^{2}}}=log{\left |x+\sqrt {x^{2}+a^{2}}\right |}+C\)
  • \(\rm \int_{0}^{x}{f(x) dx}=F(x)-F(0)\), 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.

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