Integrate with respect to x Y= (4x² +x-² +4)dx So far I have 4x/3 +x-¹/-1 + 4 + c
J Jaja New member Joined Oct 9, 2020 Messages 4 Oct 11, 2020 #1 Integrate with respect to x Y= (4x² +x-² +4)dx So far I have 4x/3 +x-¹/-1 + 4 + c
D Deleted member 4993 Guest Oct 11, 2020 #2 Jaja said: Integrate with respect to x Y= (4x² +x-² +4)dx So far I have 4x/3 +x-¹/-1 + 4 + c Click to expand... That is incorrect. You need to use parentheses () to group your operations. Use "^" to indicate exponentiation. We had pointed some of these in your previous posts. Try again. Remember: \(\displaystyle \int x^n \ \ dx = \frac{x^{n+1}}{n+1} + C\)
Jaja said: Integrate with respect to x Y= (4x² +x-² +4)dx So far I have 4x/3 +x-¹/-1 + 4 + c Click to expand... That is incorrect. You need to use parentheses () to group your operations. Use "^" to indicate exponentiation. We had pointed some of these in your previous posts. Try again. Remember: \(\displaystyle \int x^n \ \ dx = \frac{x^{n+1}}{n+1} + C\)
J JeffM Elite Member Joined Sep 14, 2012 Messages 7,874 Oct 11, 2020 #3 [MATH] \int 4 \ dx = 4???????[/MATH]
pka Elite Member Joined Jan 29, 2005 Messages 11,971 Oct 11, 2020 #4 Jaja said: Integrate with respect to x Y= (4x² +x-² +4)dx So far I have 4x/3 +x-¹/-1 + 4 + c Click to expand... \(\displaystyle\int {\left( {4{x^2} + \dfrac{1}{{{x^2}}} + 4} \right)dx = \dfrac{4}{3}{x^3} - \dfrac{1}{x} + 4x} \) SEE HERE
Jaja said: Integrate with respect to x Y= (4x² +x-² +4)dx So far I have 4x/3 +x-¹/-1 + 4 + c Click to expand... \(\displaystyle\int {\left( {4{x^2} + \dfrac{1}{{{x^2}}} + 4} \right)dx = \dfrac{4}{3}{x^3} - \dfrac{1}{x} + 4x} \) SEE HERE