Question

Intregate

(2+3x-x^4/x) dx

Answer 1

\[\large\rm \int\limits \frac{2+3x-x^4}{x}dx\quad=\int\limits \frac{2}{x}+\frac{3x}{x}-\frac{x^4}{x}dx\]Here is a good first step you can apply :)

Answer 2

I see

Answer 3

do we change 2/x to 2x

Answer 4

3x/x to 3x^2?

Answer 5

Let's actually leave the first one alone for now, that's a special integral.
But the others, yes we need to do the division,
and it looks like you're goofing it up a tad bid ^^

Answer 6

\[\large\rm \frac{3\cancel x}{\cancel x}\quad=3\]

Answer 7

2/x+3-x^3??

Answer 8

Mmm k looks good!
Let's pull the 2 off of the first integral just so it's a little clearer what's going on,\[\large\rm =\int\limits\limits 2\cdot\frac{1}{x}+3-x^3 ~dx\]

Answer 9

So recall that you can not apply your power rule to a -1 power.\[\large\rm \int\limits \frac{1}{x^1}dx\quad=\int\limits x^{-1}\quad\ne \frac{1}{-1+1}x^0\]Not allowed to do that.

Answer 10

Do you remember this one? :)
It's sort of special,\[\large\rm \int\limits \frac{1}{x}dx\quad= \ln|x|\]

Answer 11

yes i do

Answer 12

So our first term is going to change into 2 times that.

Answer 13

How about the other guys?

Answer 14

wait I'm confused..

Answer 15

how do we put 2/x in ln lxl

Answer 16

\[\large\rm =2\int\limits \frac{1}{x}dx+\int\limits3dx-\int\limits x^3 dx\]We're allowed to write this as a sum of integrals.
That will allow us to integrate term by term.

Answer 17

Make sure you're comfortable with fractions:\[\large\rm \frac{a}{b}\quad=a\cdot \frac{1}{b}\quad=\frac{1}{b}\cdot a\]

Answer 18

\[\large\rm =2\color{orangered}{\int\limits\limits \frac{1}{x}dx}+\int\limits\limits3dx-\int\limits\limits x^3 dx\]The 2 isn't doing anything fancy, he's just coming along for the ride,\[\large\rm =2\color{orangered}{\ln|x|}+\int\limits\limits3dx-\int\limits\limits x^3 dx\]

Answer 19

Hopefully you understood the step I made in the middle there,\[\large\rm \frac{2}{x}\quad=2\cdot\frac{1}{x}\]I did that to pull the 2 outside of the integral ^

Answer 20

okay i understood

Answer 21

what the next step

Answer 22

\[\large\rm =2\ln|x|+\int\limits\limits\limits3dx-\int\limits\limits\limits x^3 dx\]We took care of that weird special integral in front.
Now you need to integrate the others using power rule.

Answer 23

2 \[2\ln \left| x \right|+\frac{ 3^2 }{ 2 }-\frac{ x^4 }{ 4 }\]

Answer 24

\[\large\rm 2\ln \left| x \right|+\cancel{\frac{ 3^2 }{ 2 }}-\frac{ x^4 }{ 4 }\]Ok the last term looks good.
Let's see what's going on in the middle here...

Answer 25

Do you remember how to take this derivative?\[\large\rm (3x)'=?\]

Answer 26

I guess it was my fault for saying "power rule" on that one ;) hehe

Answer 27

No i don't remembe

Answer 28

\[\large\rm (3x)'\quad=3(x)'\quad=3(1)\]Derivative of x, with respect to x, is just 1.

Answer 29

So if \(\large\rm (3x)'=3\)

Then going backwards, \(\large\rm \int 3dx=3x\)

Answer 30

So bam, there we go,\[\large\rm =2\ln \left| x \right|+3x-\frac{ x^4 }{ 4 }+c\]our final answer.