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(4√3/5) -1 ∫ x+2. dx equals to 1 √(x 2 +2x-3) 2√3/3 – ½ log 3 2√3/3 + ½ log 3 2√3/3 – ½ log (√3+2) 2√3/3 + ½ log( √3+2)

(4√3/5) -1
       ∫     x+2.                dx     equals to
      1   √(x2+2x-3)
  1.  2√3/3 – ½ log 3
  2. 2√3/3  + ½ log 3
  3. 2√3/3 – ½ log (√3+2)
  4. 2√3/3 + ½ log( √3+2)

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1 Answers

Sushant Kumar
42 Points
5 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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