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Ask iit jee aieee pet cbse icse state board community Community Discussion Question: Integrate - Using the shortest possible no of steps
Forum Index -> Integral Calculus like the article? email it to a friend.  
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srujana (3045)

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Ans: Click here


 




 


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anandghegde (1707)

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 You want only that ans or will you accept other answers too? Huh?Huh?


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anandghegde (1707)

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 isnt this a standard kind of prob?Mr. GreenMr. Green


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srujana (3045)

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Any right ans / genuine attempt is accepted ! :)



(P.S- shd have least poss no of steps)


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srujana (3045)

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is it??



(forgive me if it is!...m a beginner at integration..!)



 


God has given you one face, and you
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~William Shakespeare

You were born an original. Don't die a copy.
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anandghegde (1707)

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 Write the integral as        



 


now proceedMr. GreenMr. Green


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srujana (3045)

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It more or less resembles my procedure....which no doubt isnt that short...




 


it gives the ans




 





 





 


 




 


(which no way resembles the given ans)


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anandghegde (1707)

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you want the answers to tally with those given by mathematica?ShockedShocked


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sandeepramesh (1245)

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There might be many ways of integrating the expression! By no means need all the answers tally! Mr. Green Mr. Green

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srujana (3045)

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hey ppl...!

atleast try completing the sum...

(note the word completing..!)

n as i have said...m luking fr any "simple trick" which these kind of qs usually have to crack them n leave ppl surprised !

God has given you one face, and you
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You were born an original. Don't die a copy.
~John Mason
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anandghegde (1707)

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 no simple trick......this is a standard prob Mr. GreenMr. Green


of this type



 


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sandeepramesh (1245)

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maybe it helps if you know what on integrating gives sinh^{-1} function thats all 

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sboosy (3011)

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\int (x^2+2x+3)\sqrt{x^2+x+1} \,dx \\ \\ = \int x^2\sqrt{x^2+x+1} \,dx+ \int (2x+1)\sqrt{x^2+x+1} \,dx + 2\int \sqrt{x^2+x+1} \,dx \\ \\ \int \sqrt{x^2+x+1} \,dx = \int \sqrt{\left(x+\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2} \,dx \\ \\ \mbox{Apply formula} \ \int \sqrt{a^2+x^2} \,dx \ \mbox{to get the answer.Let it be K} \\ \\ \int (2x+1)\sqrt{x^2+x+1} \,dx \ \mbox{Put} \ x^2+x+1=t^2,(2x+1)dx = 2tdt \\ \\ \int x^2\sqrt{x^2+x+1} \,dx = \int x^2 d(K) = (x^2\times K) - \int K(2x)\,dx \ \mbox{where} \\ \\ K = \frac{x\sqrt{a^2+x^2}}{2}+\frac{a^2 \log(x+\sqrt{a^2+x^2})}{2}\\ \\\mbox{From here it is easy}


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