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Question-214072




Question Number 214072 by Mwamba last updated on 26/Nov/24
Answered by TonyCWX08 last updated on 27/Nov/24
Answered by TonyCWX08 last updated on 27/Nov/24
Consider two points, P and Q.  P = (x,y) and Q=(x+δx,y+δy)  Let s = length of arc from starting point  δs = Length of arc PQ    Since δs is very small, we can approximate arc PQ to a straight line.  By Pythagoras′s theorem,  (δx)^2 +(δy)^2 =(δs)^2     Dividing both sides by (δx)^2 , we get  1+(((δy)/(δx)))^2 =(((δs)/(δx)))^2     As δx→0, this gives  1+((dy/dx))^2 =((ds/dx))^2     Applying Square Root on both sides,  (ds/dx)=(√(1+((dy/dx))^2 ))    Therefore, we have  s=∫(√(1+((dy/dx))^2 ))dx or ∫(√(1+(f′(x))^2 ))dx
Considertwopoints,PandQ.P=(x,y)andQ=(x+δx,y+δy)Lets=lengthofarcfromstartingpointδs=LengthofarcPQSinceδsisverysmall,wecanapproximatearcPQtoastraightline.ByPythagorasstheorem,(δx)2+(δy)2=(δs)2Dividingbothsidesby(δx)2,weget1+(δyδx)2=(δsδx)2Asδx0,thisgives1+(dydx)2=(dsdx)2ApplyingSquareRootonbothsides,dsdx=1+(dydx)2Therefore,wehaves=1+(dydx)2dxor1+(f(x))2dx
Answered by TonyCWX08 last updated on 27/Nov/24
∫(dx/(x^4 +5x^2 +6))  =∫(dx/((x^2 +3)(x^2 +2)))  =∫(−(1/(x^2 +3))+(1/(x^2 +2)))dx  =−∫((1/(x^2 +3)))dx+∫((1/(x^2 +2)))dx  =−(1/( (√3)))(tan^(−1) ((x/( (√3)))))+(1/( (√2)))(tan^(−1) ((x/( (√2)))))+C
dxx4+5x2+6=dx(x2+3)(x2+2)=(1x2+3+1x2+2)dx=(1x2+3)dx+(1x2+2)dx=13(tan1(x3))+12(tan1(x2))+C
Answered by lepuissantcedricjunior last updated on 28/Nov/24
b)((𝛑(3(√2)−2(√3)))/(12))
\boldsymbolb)\boldsymbolπ(3223)12b)π(3223)12

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