1. Try u=e^(7x)+3
A picture is attached...both involve substitution but what I don't understand is for the first one I know u would be e^7x and du would then be e^7x but where does that leave me? How do I decide what becomes u and what remains e^7x in the function?
For the second I know u would be x^3 and du would be 3x^2 so you'd have 1/3 the integral of u times the square root of...SOMETHING. Now would the x^2 inside the sq root just disappear leaving with 42 or what?
Don't forget to use the chain rule:
The power rule for integration is Obviously, this cannot work for an exponent of -1, because then there would be a division by zero. Instead, you have to use the log rule, which Jameson listed above.Alright so then my next step would be the integral of 1/u and the antiderivative of that would be...what? Because 1/u is just u^-1 so wouldn't the antiderivative just be u^0 or 1?
Let ThenFor the second I know u would be x^3 and du would be 3x^2 so you'd have 1/3 the integral of u times the square root of...SOMETHING. Now would the x^2 inside the sq root just disappear leaving with 42 or what?
Substitution gives
But here's the kicker: means that So we can substitute for to get
OK so going from what Reckoner said I did
1/2((5u^5/2 divided by 2)-3u^3/2 divided by 2)) and the answer is incorrect. I put in x^2+42 for every u you see in the problem and my final (incorrect) response, did it 5 times with a calculator and got the same answer so I must be doing something wrong with these antiderivatives.