Question

Im doing integration by parts and need assistance. Im not sure if I should use the chain rule or not.

Answer 1

\[\int\limits( x^2+1) ^2 dx\]

Answer 2

i expanded but now Im stuck.

Answer 3

\[\int\limits x^4 +x^2+1\]

Answer 4

After expanding, it should be\[\int\limits\limits( x^2+1) ^2 dx = \int\limits [x^4 + 2x^2 + 1]dx\]

Answer 5

\[\int\limits\] of 1?

Answer 6

Have you learned derivatives? What is the antiderivative of 1? The derivative of x is 1...

Answer 7

Thats the part I know. I know the derivative of x is 1. But I cant seem to come up with the anti derivative because I ended up with (1^2)/2

Answer 8

1 is another way of saying \[x^0 \]

\[\int\limits 1 dx \rightarrow \int\limits x^0 dx = {x^1 \over 1} = x\]

Answer 9

anti derivative of x^m = x^(m+1) / (m+1)

Answer 10

So the antideriative comes out to x?

Answer 11

Yeah. Anti-derivative of 1 comes out as x when you are integrating with respect to x (that's what the dx stands for).

Answer 12

So back to the problem, Im stuck because after I expanded and did the antiderivative, it looks nothing like my choices.

Answer 13

It's because you've got a function squared, therefore the easiest way to tackle this is to multiply out the brackets. Or you could use u-substitution.\[\int\limits (x^2 + 1)^2 dx = \int\limits (x^2 +1)(x^2 + 1) dx = \int\limits (x^4 +2x^2 + 1) dx = {x^5 \over 5} + {x^3 \over 3} + {x^1 \over 1} + c\]

Answer 14

I have all that but where do I go from there?

Answer 15

what are your choices?

Answer 16

btw, there's a typo in Will's answer:
\[\frac{x^5}{5}+\frac{2x^3}{3}+x+c \]

Answer 17

Sorry, It's late! Cheers :P

Answer 18

A: (1/3)(x^2+1)^3+c B: (1/6x)(x^2+1)^3+c C: (2x/3)(x^2 +1)^3 +c D:(x^3/3 +x)^2 +c E:(1/5)x^5 +(2/3)x^3+x+c

Answer 19

doesn't E look familiar?

Answer 20

hEhEhEhEhEhEhEh(it's E)hEhEhEhE

Answer 21

hahahahahaahaha Sorry, I've been doing a calculus packet for hours now so my brain is frayed!

Answer 22

time for a break

Answer 23

I think so. Ihavent had a break since I started and that was around 2.

Answer 24

Send it to me, I'm bored :P

Answer 25

btw you could do integration by parts, but it is more work than just expanding the square.

u = (x^2+1)^2 du = 2(x^2+1) 2x dx = 4x(x^2+1) = 4x^3 + 4x dx
dv = dx v= x
uv - integral v du = x(x^2+1)^2 - integral x(4x^3 +4x) dx
the integral is 4(integral x^4 +x^2)= (4/5)x^5 + (4/3)x^3
and we have
x(x^2+1)^2 - (4/5)x^5 - (4/3)x^3
now you still have to expand the square just to simplify this mess, but it comes out as E

Answer 26

+ constant

Answer 27

Yes, thank you. That one looks a little more familiar. But the other way was a lot easier.