(7^3square root of x^4 - 2/square root of x) dx du=x^4...
A Ashley5 New member Joined Nov 3, 2007 Messages 14 Nov 20, 2007 #1 (7^3square root of x^4 - 2/square root of x) dx du=x^4 dx=4x^3 Could someone help me finish this problem. Thank you
(7^3square root of x^4 - 2/square root of x) dx du=x^4 dx=4x^3 Could someone help me finish this problem. Thank you
tkhunny Moderator Staff member Joined Apr 12, 2005 Messages 11,339 Nov 20, 2007 #2 73x4−2x\displaystyle \frac{7^{3\sqrt{x^{4}-2}}}{\sqrt{x}}x73x4−2?
A Ashley5 New member Joined Nov 3, 2007 Messages 14 Nov 20, 2007 #3 yes, sorry I didn't know how to type it.
tkhunny Moderator Staff member Joined Apr 12, 2005 Messages 11,339 Nov 21, 2007 #4 You're kidding!? Why do you think you need to find this anitderivative? Can you supply the complete problem statement?
You're kidding!? Why do you think you need to find this anitderivative? Can you supply the complete problem statement?
D Deleted member 4993 Guest Nov 21, 2007 #5 tkhunny said: 73x4−2x\displaystyle \frac{7^{3\sqrt{x^{4}-2}}}{\sqrt{x}}x73x4−2? Click to expand... or is it: 73⋅ x4−2x\displaystyle \frac{7^{3}\cdot\ \sqrt{x^{4}-2}}{\sqrt{x}}x73⋅ x4−2
tkhunny said: 73x4−2x\displaystyle \frac{7^{3\sqrt{x^{4}-2}}}{\sqrt{x}}x73x4−2? Click to expand... or is it: 73⋅ x4−2x\displaystyle \frac{7^{3}\cdot\ \sqrt{x^{4}-2}}{\sqrt{x}}x73⋅ x4−2
G galactus Super Moderator Staff member Joined Sep 28, 2005 Messages 7,216 Nov 21, 2007 #6 Without proper grouping symbols(or better yet, LaTex) it's difficult for us to tell what you mean. 73∫x4−2xdx\displaystyle 7^{3}\int\frac{\sqrt{x^{4}-2}}{\sqrt{x}}dx73∫xx4−2dx is not easily integrated. Perhaps you mean: \(\displaystyle 7^{3}\int[\sqrt{x^{4}}-\frac{2}{\sqrt{x}}}]dx\)?.
Without proper grouping symbols(or better yet, LaTex) it's difficult for us to tell what you mean. 73∫x4−2xdx\displaystyle 7^{3}\int\frac{\sqrt{x^{4}-2}}{\sqrt{x}}dx73∫xx4−2dx is not easily integrated. Perhaps you mean: \(\displaystyle 7^{3}\int[\sqrt{x^{4}}-\frac{2}{\sqrt{x}}}]dx\)?.
