Question

integral 1/ √x^2-x-1 dx

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Answer 1

Solution

We have

$\to \sf \int \dfrac{1}{ \sqrt{ {x}^{2} - x - 1} } dx$

For Solving this type of question we have to convert into prefect square

Now Take

$\sf \to \: {x}^{2} - x - 1 = 0$

$\sf \to {x}^{2} - x = 1$

$\sf \to \: {x}^{2} - x + \bigg( \dfrac{1}{2} \bigg) ^{2} - \bigg( \dfrac{1}{2} \bigg) ^{2} = 1$

$\sf \to \: \bigg(x - \dfrac{1}{2} \bigg) ^{2} - \bigg( \dfrac{1}{2} \bigg)^{2} = 1$

$\sf \to \: \bigg(x - \dfrac{1}{2} \bigg) ^{2} = 1 + \dfrac{1}{4}$

$\sf \to \: \bigg(x - \dfrac{1}{2} \bigg) ^{2} = \dfrac{5}{4}$

$\sf \to \bigg(x - \dfrac{1}{2} \bigg) ^{2} - \bigg( \dfrac{ \sqrt{5} }{2} \bigg) ^{2}$

Now we can write as

$\sf \to \: {x}^{2} - x - 1 = \bigg(x - \dfrac{1}{2} \bigg) ^{2} - \bigg( \dfrac{ \sqrt{5} }{2} \bigg) ^{2}$

$\to \sf \int \dfrac{1}{ \sqrt{ \bigg(x - \dfrac{1}{2} \bigg) ^{2} - \bigg( \dfrac{ \sqrt{5} }{2} \bigg) ^{2}} } dx$

Now using Loving Intergrals

$\sf \to \: \int \: \dfrac{dx}{ \sqrt{ {x}^{2} - {a}^{2} } } = ln(x + \sqrt{ {x}^{2} - a^{2}} ) + \: c$

By comparing we get

$\sf \to \: x = x - \dfrac{1}{2} \: \: \: and \: \: a \: = \dfrac{ \sqrt{5} }{2}$

put the value we get

$\sf \to ln \bigg \{ \bigg(x - \dfrac{1}{2} \bigg) + \sqrt{ \bigg(x - \dfrac{1}{2} \bigg) ^{2} - \bigg( \dfrac{ \sqrt{5} }{2} \bigg)^{2} } \bigg \} + c$

Answer

$\sf \to ln \bigg \{ \bigg(x - \dfrac{1}{2} \bigg) + \sqrt{ \bigg(x - \dfrac{1}{2} \bigg) ^{2} - \bigg( \dfrac{ \sqrt{5} }{2} \bigg)^{2} } \bigg \} + c$