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(4√3/5) -1∫x+2.dxequals to1√(x2+2x-3)2√3/3 – ½ log 3\t2√3/3+ ½ log 3\t2√3/3 – ½ log (√3+2)\t2√3/3 + ½ log( √3+2)

Technical GURUJI , 6 Years ago
Grade
anser 1 Answers
Sushant Kumar

Last Activity: 6 Years ago

first we multiply and divide by 2 to get 
1/2\int (2x+4)/\sqrt{x^2+2x-3}     (not putting the limits now just for simplicity)
 
this wil become 1/2[\int (2x+2)/\sqrt{x^2+2x-3} +2*\int \frac{1}{\sqrt{x^2+2x-3}}] 
 
now take x^2+2x-3=t
change the integration limits
and the first integral will become \int \frac{1}{\sqrt{t}}  and the other integral is of the form \int \frac{1}{\sqrt{(x+1)^{2}-2^{2}}}
 
hope this will help

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