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Question: The \(\int {\dfrac{{\sqrt {x - 1} }}{{x\sqrt {x + 1} }}dx} \) is equal to $(a)\ln |x - \sqrt {{x^2...

The ∫x−1xx+1dx\int {\dfrac{{\sqrt {x - 1} }}{{x\sqrt {x + 1} }}dx} is equal to
(a)\ln |x - \sqrt {{x^2} - 1} | - {\tan ^{ - 1}}x + C \\\ (b)\ln |x + \sqrt {{x^2} - 1} | - {\tan ^{ - 1}}x + C \\\ (c)\ln |x - \sqrt {{x^2} - 1} | - {\sec ^{ - 1}}x + C \\\ (d)\ln |x + \sqrt {{x^2} - 1} | - {\sec ^{ - 1}} + C \\\

Explanation

Solution

The very first step we need to do is rationalize the given form and therefore, we will multiply and divide both the numerator and denominator with x−1\sqrt {x - 1} . After this step we will split the resultant answer into 2 parts and solve them individually. One part of the answer when integrated will give us the answer ln⁡(x+x2−1)\ln \left( {x + \sqrt {{x^2} - 1} } \right)and the other one will give a direct formula answer and that is sec⁡−1x+C{\sec ^{ - 1}}x + C.When we combine these two we will know that option D is our final answer.

Complete step by step answer:
The first step to solve this problem will be to multiply both the numerator and denominator with x−1\sqrt {x - 1} . So,
∫x−1xx+1dx=∫x−1×x−1xx+1×x−1dx ⇒∫x−1xx+1dx=∫(x−1)2xx2−1dx ....................[∵(a+b)(a−b)=a2−b2] ⇒∫x−1xx+1dx=∫x−1xx2−1dx ⇒∫1x2−1dx−∫1xx2−1dx \int {\dfrac{{\sqrt {x - 1} }}{{x\sqrt {x + 1} }}} dx = \int {\dfrac{{\sqrt {x - 1} \times \sqrt {x - 1} }}{{x\sqrt {x + 1} \times \sqrt {x - 1} }}} dx \\\ \Rightarrow\int {\dfrac{{\sqrt {x - 1} }}{{x\sqrt {x + 1} }}} dx= \int {\dfrac{{{{\left( {\sqrt {x - 1} } \right)}^2}}}{{x\sqrt {{x^2} - 1} }}} dx\,....................\left[ {\because \left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}} \right] \\\ \Rightarrow\int {\dfrac{{\sqrt {x - 1} }}{{x\sqrt {x + 1} }}} dx= \int {\dfrac{{x - 1}}{{x\sqrt {{x^2} - 1} }}dx} \\\ \Rightarrow\int {\dfrac{1}{{\sqrt {{x^2} - 1} }}dx - \int {\dfrac{1}{{x\sqrt {{x^2} - 1} }}} } dx \\\
Now, we will consider each part of the sum separately to make it easier. Let’s
a=1x2−1a = \dfrac{1}{{\sqrt {{x^2} - 1} }}and b=1xx2−1b = \dfrac{1}{{x\sqrt {{x^2} - 1} }}
Hence, ∫x−1xx+1dx=∫a−∫b\int {\dfrac{{\sqrt {x - 1} }}{{x\sqrt {x + 1} }}dx = } \int a - \int b
Now, solving the “a” part first we will get:-
a=∫1x2−1dx ⇒a=ln⁡(x+x2−1) a = \int {\dfrac{1}{{\sqrt {{x^2} - 1} }}} dx \\\ \Rightarrow a= \ln \left( {x + \sqrt {{x^2} - 1} } \right) \\\
Now, solving the “b part
b=∫1xx2−1dxb = \int {\dfrac{1}{{x\sqrt {{x^2} - 1} }}} dx
Now, we know that,
ddxsec⁡−1(x)=1xx2−1\dfrac{d}{{dx}}{\sec ^{ - 1}}(x) = \dfrac{1}{{x\sqrt {{x^2} - 1} }}
Therefore,
∫1xx2−1dx=sec⁡−1x+C\int {\dfrac{1}{{x\sqrt {{x^2} - 1} }}dx = {{\sec }^{ - 1}}x + C}
Where, C is a constant
Hence, when we combine both the results of a and B we will get
∫x−1xx+1dx=ln⁡(x+x2−1)−sec⁡−1+C\int {\dfrac{{\sqrt {x - 1} }}{{x\sqrt {x + 1} }}dx = \ln \left( {x + \sqrt {{x^2} - 1} } \right)} - {\sec ^{ - 1}} + C

Therefore, option D is the correct choice.

Note: The first step to rationalize the given form was done just to simplify our calculation. You need to think out of the box sometimes to solve questions. After that step, almost every step needed deep calculations and strong command on integration. Therefore, remember every formula and method of integration. The part where we solved a given integral directly with sec⁡−1x+C{\sec ^{ - 1}}x + C can be done only if you remember all the required formulas.