1/[(6x+.01x^2)^1/2] , with respect to x. Thank you in advance.
S Stewart New member Joined Sep 23, 2006 Messages 2 Sep 25, 2006 #1 1/[(6x+.01x^2)^1/2] , with respect to x. Thank you in advance.
S soroban Elite Member Joined Jan 28, 2005 Messages 5,586 Sep 25, 2006 #2 Hello, Stewart! \(\displaystyle \L\int \frac{dx}{\sqrt{6x\,+\,0.01x^2}}\) Click to expand... Complete the square . . . 6x + 0.01x2 = 0.01(x2 + 600x)\displaystyle 6x\,+\,0.01x^2 \;=\;0.01\left(x^2\,+\,600x\right)6x+0.01x2=0.01(x2+600x) . . . . . . . . .= 0.01(x2 + 600x + 3002 − 3002)\displaystyle = \;0.01\left(x^2\,+\,600x\,+\,300^2\,-\,300^2\right)=0.01(x2+600x+3002−3002) . . . . . . . . .= 0.01[(x + 300)2 − 3002]\displaystyle = \;0.01\left[(x\,+\,300)^2\,-\,300^2\right]=0.01[(x+300)2−3002] 6x + 0.01x2 = 0.1(x + 300)2 − 3002\displaystyle \sqrt{6x\,+\,0.01x^2}\;=\;0.1\sqrt{(x\,+\,300)^2\,-\,300^2}6x+0.01x2=0.1(x+300)2−3002 The integral becomes: \(\displaystyle \:10\L\int\frac{dx}{\sqrt{(x\,+\,300)^2\,-\,300^2}}\) Use the substitution: x + 300 = 300secθ\displaystyle \,x\,+\,300\:=\:300\sec\thetax+300=300secθ
Hello, Stewart! \(\displaystyle \L\int \frac{dx}{\sqrt{6x\,+\,0.01x^2}}\) Click to expand... Complete the square . . . 6x + 0.01x2 = 0.01(x2 + 600x)\displaystyle 6x\,+\,0.01x^2 \;=\;0.01\left(x^2\,+\,600x\right)6x+0.01x2=0.01(x2+600x) . . . . . . . . .= 0.01(x2 + 600x + 3002 − 3002)\displaystyle = \;0.01\left(x^2\,+\,600x\,+\,300^2\,-\,300^2\right)=0.01(x2+600x+3002−3002) . . . . . . . . .= 0.01[(x + 300)2 − 3002]\displaystyle = \;0.01\left[(x\,+\,300)^2\,-\,300^2\right]=0.01[(x+300)2−3002] 6x + 0.01x2 = 0.1(x + 300)2 − 3002\displaystyle \sqrt{6x\,+\,0.01x^2}\;=\;0.1\sqrt{(x\,+\,300)^2\,-\,300^2}6x+0.01x2=0.1(x+300)2−3002 The integral becomes: \(\displaystyle \:10\L\int\frac{dx}{\sqrt{(x\,+\,300)^2\,-\,300^2}}\) Use the substitution: x + 300 = 300secθ\displaystyle \,x\,+\,300\:=\:300\sec\thetax+300=300secθ
S Stewart New member Joined Sep 23, 2006 Messages 2 Sep 26, 2006 #3 Thanx. Soroban. Your info was very helpful.
