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 . . . \(\displaystyle 6x\,+\,0.01x^2 \;=\;0.01\left(x^2\,+\,600x\right)\) . . . . . . . . .\(\displaystyle = \;0.01\left(x^2\,+\,600x\,+\,300^2\,-\,300^2\right)\) . . . . . . . . .\(\displaystyle = \;0.01\left[(x\,+\,300)^2\,-\,300^2\right]\) \(\displaystyle \sqrt{6x\,+\,0.01x^2}\;=\;0.1\sqrt{(x\,+\,300)^2\,-\,300^2}\) The integral becomes: \(\displaystyle \:10\L\int\frac{dx}{\sqrt{(x\,+\,300)^2\,-\,300^2}}\) Use the substitution: \(\displaystyle \,x\,+\,300\:=\:300\sec\theta\)
Hello, Stewart! \(\displaystyle \L\int \frac{dx}{\sqrt{6x\,+\,0.01x^2}}\) Click to expand... Complete the square . . . \(\displaystyle 6x\,+\,0.01x^2 \;=\;0.01\left(x^2\,+\,600x\right)\) . . . . . . . . .\(\displaystyle = \;0.01\left(x^2\,+\,600x\,+\,300^2\,-\,300^2\right)\) . . . . . . . . .\(\displaystyle = \;0.01\left[(x\,+\,300)^2\,-\,300^2\right]\) \(\displaystyle \sqrt{6x\,+\,0.01x^2}\;=\;0.1\sqrt{(x\,+\,300)^2\,-\,300^2}\) The integral becomes: \(\displaystyle \:10\L\int\frac{dx}{\sqrt{(x\,+\,300)^2\,-\,300^2}}\) Use the substitution: \(\displaystyle \,x\,+\,300\:=\:300\sec\theta\)
S Stewart New member Joined Sep 23, 2006 Messages 2 Sep 26, 2006 #3 Thanx. Soroban. Your info was very helpful.
