Question

Find the value of the \(\rm \int_{5}^{6} \frac{dx}{\sqrt {x^{2}-16}}\)
1. \(\rm log\frac{ {|6-\sqrt {20}}|}{{3}}\)
2. \(\rm log\frac{ {|6+\sqrt {20}}|}{{5}}\)
3. \(\rm log\frac{ {|3+\sqrt {20}}|}{{3}}\)
4. \(\rm log\frac{ {|6+\sqrt {20}}|}{{8}}\)

Answer 1

Correct Answer - Option 4 : \(\rm log\frac{ {|6+\sqrt {20}}|}{{8}}\)

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: \(\rm \int_{5}^{6} \frac{dx}{\sqrt {x^{2}-16}}\)

Using the formula, \(\rm \int\frac{dx}{\sqrt {x^{2}-a^{2}}}=log{\left |x+\sqrt {x^{2}-a^{2}}\right |}+C\)

\(\Rightarrow \rm \int_{5}^{6} \frac{dx}{\sqrt {x^{2}-16}}= \int_{5}^{6} \frac{dx}{\sqrt {x^{2}-4^{2}}}=\left [ log{\left |x+\sqrt {x^{2}-16}\right |} \right ]_{5}^{6}\)

Now, Substitute the limit to evaluate the value.

​\(\Rightarrow \rm \int_{5}^{6} \frac{dx}{\sqrt {x^{2}-16}}=log{|6+\sqrt {6^{2}-16}}|-log{|5+\sqrt {5^{2}-16}}|=log{|6+\sqrt {20}}|-log{|8}|\)​

It is known that log a - log b = log (a / b)

\(\Rightarrow \rm \int_{5}^{6} \frac{dx}{\sqrt {x^{2}-16}}=log\frac{ {|6+\sqrt {20}}|}{{8}}\)

Hence, the correct answer is option 4.