What is the derivative of ln(x)? Work backwards to know when the integral will be ln∣x∣∫52−x3dx....Missing parentheses?
→52x−?+C :?: How to I use the ln integral formula.
How are you getting that the derivative of ln(3x) is 3/x?∫52−x3
52x−ln∣3x∣+C - The computer keeps saying wrong. :?:
How are you getting that the derivative of ln(3x) is 3/x?![]()
A feeling a lot of us have regarding, well, you know.I give up!!
How are you getting that the derivative of ln(3x) is 3/x?
Repeating the fact that you'd somehow arrived at a wrong result does not explain how you arrived at that wrong result. Please show your work, especially when it is asked that you do. Thank you!dxdln3x→x3
Surely you learned, long ago, that ∫Af(x)dx=A∫f(x)dx for any constant, A?∫52−x3dx
→52x−?+C :?: How to I use the ln integral formula.
I think "mad" is the wrong word here. More like frustrated. You continually post problems where you show your work (good thing), then when we critique your work, whether it's bad algebra or incorrect usage of formulas, substitution principles, etc., you tend to not correct these mistakes in your next thread post. It drives me, and I'm sure many others here, crazy to where we want to bang our heads on the wall.Sorry if i made anyone mad on here.
.Sorry if i made anyone mad on here. I see what the answer is. I will do some work and see if I can understand the mechanics behind it. Also, sorry i hadn't replied sooner, but I was busy doing homework all day.
∫x1dx→ln∣u∣+C
So considering ∫x3dx then what isu?
Continuing with the theme expressed in post above, why would you want to use substitution here?
Substitution is used when the original expression is complicated.
Here you should know (and I have indicated before) that ∫x1dx=ln(∣x∣)+C .......... use that!!
u must be x3 but how did it get there?
Perhaps it can be understood as
∫3x−1→03x0 so that becomes ln∣x3∣+C