Question

antiderivative of (1/x^2) dx?

Answer 1

Rewrite it using negative exponents, then apply the power rule for integration! :) It shouldn't be too bad :D Lemme know if any of that is a little confusing.

Answer 2

i have got x^-2 dx = x^-1/-1 + c, is that right?

Answer 3

Yes looks good ! ^^

Answer 4

Last step might be to rewrite it without the fraction part though, so it looks a little cleaner :)\[-x^{-1}+c\]

Answer 5

i have 7 more antiderivative problems...I have attempted all of them I just need someone to verify. would you be willing to help me?

Answer 6

sure, just post your next one in here if you like. I'll try to help if I have time :D But I'm sure someone can get to you if I don't! :D

Answer 7

ok next one is similar to first

antiderivative of (1/x) dx

Answer 8

Unfortunately, the power rule for integration doesn't apply to x^-1.
Remember what this special one gives you? :O

Answer 9

what exactly is the power rule of integration btw

Answer 10

\[\huge \int\limits_{}^{}x^n dx=\frac{ x^{n+1} }{ n+1 }+c\]

Answer 11

You increase the exponent by one, then divide by the NEW power.

Answer 12

how does this explain the -1 in the denominator of the first problem

Answer 13

\[\huge \int\limits_{}^{}x^{-2}dx=\frac{ x^{-2+1} }{ -2+1 }+c=\frac{ x^{-1} }{ -1 }+c\]

Answer 14

i see how now. thanks

Answer 15

what about the second one, what would you do since you cant use power rule of integration

Answer 16

It's one that you'll end up memorizing.
If you don't know what it is, you need to go back to differentiation.
\[\huge (\ln x)' = \frac{ 1 }{ x }\]\[\huge \int\limits_{}^{}\frac{ 1 }{ x }dx=?\]

Answer 17

ln[x] + c?

Answer 18

yes good c: Now there is something going on with the domain when you take this integral. So when you deal with DEFINITE integrals, you want to remember that it's actually producing \[\huge \ln |x| + c\]
Absolute x.

Answer 19

But that's not something we'll really need to worry about for indefinite integrals :D

Answer 20

ok thanks for that one

next one is
antiderivative of x^3 (x^4 + 1) ^2 dx

Answer 21

So this is where you get an introduction to U substitutions. Hmm

Answer 22

See how we have an x^4, and an x^3?

We can use the fact that they're one power apart to think of it as (Something)... and the derivative of (Something)

Let's apply a u substitution.
Let \[\huge u=x^4+1\]
Taking the derivative of both sides with respect to x...
\[\huge \frac{ du }{ dx }=3x^3\]

Answer 23

wouldnt the derivative be 4x^3

Answer 24

Yes thank you XD sorry about that.

Answer 25

ok no problem

Answer 26

From here, there is a process that allows you to rewrite the dx on the other side next to the 4x^3. Just simply think of it as multiplication. That's not entirely accurate, but for our purposes it is fine to think of it that way :D
\[\huge du=4x^3dx\]

Answer 27

for this part i wrote it as 1/4du = x^3 dx

Answer 28

K good, you're thinking ahead c: