想請問幾題不定積分與不定積分

1. w1 = ∫(0~5) √x÷[√x+√(5-x)] dx= ∫√x[√x-√(5-x)]dx/(x-5+x)= ∫[x - √(5x-x^2)]dx / (2x-5)= ∫xdx/(2x-5) - ∫√[25/4 - (x-5/2)^2]dx/(2x-5)= ∫[1/2 + (5/2)/(2x-5)]dx - 0.5∫√[25/4 - (x-5/2)^2]dx/(x-5/2)= x + 2.5∫dx/(2x-5) - 0.5∫√(25/4-y^2)dy/y;;;y=x-5/2= x + (5/4)ln(2x-5) - 0.5∫√(a^2-y^2)d(y/a)/(y/a);;;a=5/2= x + (5/4)ln(2x-5) - 0.5*A
A =∫√(a^2-y^2)d(y/a)/(y/a)= a∫√[1-(y/a)^2]d(y/a)/(y/a)∫∫∫∫∫∫∫∫∫= a∫√(1-sin^2Q);;;y=a*sinQ => dy=a*cosQdQ= a∫cosQcosQdQ/sinQ= a∫(1-sin^2Q)dQ/sinQ= a∫cscQdQ - a∫sinQdQ= a*ln(cscQ - cotQ) + a*cosQ= a*ln[a/y - √(a^2-y^2)/y] + a√(a^2-y^2)/a= a{ln[a/(x-a) - √(a^2-(x-a)^2)] + √[a^2-(x-a)^2]};;;x=0~5= a{ln[a/(5-a)-√(a^2-(5-a)^2)] - ln[a/a - √(a^2-a^2)]+ √[a^2-(5-a)^2] - √(a^2 - a^2)}= a{ln[(a/a)-0] - ln(1-0) + √0 - √0}= a*[ln(1) - ln(1)]= 0w1 = x + (5/4)ln(2x-5);;;x=0~5= (5 - 0) + (5/4)*ln|5/(-5)| = 5 + (5/4)*ln(1)= 5= answer
2. w2 = ∫1÷ ( x^1/2+x^1/3 ) dxx = y^6 ==> dx = 6y^5*dy w2 = ∫6y^5*dy / (y^3 + y^2)= ∫6y^3*dy / (y + 1)= 6∫(y^2 - y + 1)dy - 6∫dy/(y + 1)= 6(y^3/3 - y^2/2 + y) - 6*ln(y+1) + C= 2y^3 - 3y^2 + 6y - 6*ln(y+1) + C= 2√x - 3∛x + 6x^(1/6) - 6*ln[x^(1/6)+1] + C= answer

3.∫[(e^2x) -1] ÷ [(e^2x) +1] dx= ∫[e^x - e^(-x)]dx / [e^x + e^(-x)]= ∫sinh(x)dx/cosh(x)= ∫d[cosh(x)] / cosh(x)= ln[cosh(x)] + C= answer

https://www.wolframalpha.com/input/?i=%E2%88%AB%201%C3%B7%20(%20x%5E1%2F2%2Bx%5E1%2F3%20)%20dx&t=crmtb01


let y = x^(1/6)

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