Question

evaluate integral x+1/root x^2-x+1
​

Answer 1

Step-by-step explanation:

We have,

$I = \int \frac{x + 1}{ \sqrt{ {x}^{2} - x + 1 }} dx$

$\implies \: I = \frac{1}{2} \int \frac{2(x + 1)}{ \sqrt{{x}^{2} - x + 1 }} dx$

$\implies \: I = \frac{1}{2} \int \frac{2x + 2}{ \sqrt{ {x}^{2} - x + 1 }} dx$

$\implies \: I = \frac{1}{2} \int \frac{2x - 1 + 3}{ \sqrt{{x}^{2} - x + 1} } dx$

$\implies \: I = \frac{1}{2} \int \frac{2x - 1}{ \sqrt{ {x}^{2} - x + 1 }} dx + \frac{1}{2} \int \frac{3}{ \sqrt{ {x}^{2} - x + 1 } }$

$\implies \: I = I_{1} + I_{2}$

$\tt\:\bold{\purple{Solving\:\: I_{1} :}}$

$I_{1} = \frac{1}{2} \int \frac{2x - 1}{ \sqrt{ {x}^{2} - x + 1 } }$

Let$x^2-x+1=t^2$

$\implies\:(2x-1)dx=2t\:dt$

So,

$I_{1} = \frac{1}{2} \int \frac{2t \: dt}{ \sqrt{ {t}^{2} } }$

$\implies \: I_{1} = \int \frac{t \: dt}{t }$

$\implies \: I_{1} = \int dt$

$\implies \: I_{1} = t + c_{1}$

$\implies \: I_{1} = \sqrt{ {x}^{2} - x + 1 } + c_{1}$

$\tt\:\bold{\purple{Solving\:\: I_{2} :}}$

$I_{2} = \frac{1}{2} \int \frac{3}{ \sqrt{ {x}^{2} - x + 1 } } dx$

$\implies \: I_{2} = \frac{3}{2} \int \frac{1}{ \sqrt{ {x}^{2} - x + 1 } } dx$

$\implies \: I_{2} = \frac{3}{2} \int \frac{1}{ \sqrt{ {x}^{2} -2. (\frac{1}{2} ) x + \frac{1}{4} + \frac{3}{4} } } dx$

$\implies \: I_{2} = \frac{3}{2} \int \frac{1}{ \sqrt{ {x}^{2} -2. (\frac{1}{2} ) x + ( \frac{1}{2}) ^{2} + ( \frac{ \sqrt{3} }{2} } ) ^{2} } dx$

$\implies \: I_{2} = \frac{3}{2} \int \frac{1}{ \sqrt{ ({x - \frac{1}{2} })^{2} + ( \frac{ \sqrt{3} }{2} } ) ^{2} } dx$

$\implies \: I_{2} = \ln \bigg| \bigg(x - \frac{1}{2} \bigg) + \sqrt{ {x}^{2} - x + 1} \bigg| + c_{2}$

So, required integral

$\implies \: I= \sqrt{ {x}^{2} - x + 1} + c _{1} + \ln \bigg| \bigg(x - \frac{1}{2} \bigg) + \sqrt{ {x}^{2} - x + 1} \bigg| + c_{2}$

$\implies \: I= \sqrt{ {x}^{2} - x + 1} + \ln \bigg| \bigg(x - \frac{1}{2} \bigg) + \sqrt{ {x}^{2} - x + 1} \bigg| + C$

Where, $C=c_{1}+c_{2}$