Stuck on this problem. Please help!
1+u^4/u du
I assume you meant ∫(1+x^4)/x dx
= ∫( 1/x + x^3) dx
= lnx + (1/4)x^4 + c
oops, change my x's to u's
Thank you so very much for your help....
To solve the problem, we need to simplify the expression and integrate it.
Let's break down the expression and simplify it step by step:
1. Start by simplifying the numerator: u^4/u = u^(4-1) = u^3.
2. Our expression becomes: (1 + u^3) du.
3. Now, we can integrate the function. Recall that the integral of u^n du is (u^(n+1))/(n+1) + C, where C is the constant of integration.
4. Integrate (1 + u^3) du as follows:
- Integrating 1 du gives us u.
- Integrating u^3 du gives us (u^(3+1))/(3+1) = (u^4)/4.
5. Putting it together, the final integral of (1 + u^3) du is:
Integral [(1 + u^3) du] = (u + u^4/4) + C, where C is the constant of integration.
So, the solution to the problem is (u + u^4/4) + C, where C is the constant of integration.