Question

i need some help finding the antiderivative of 5x(x^2+1)^4

Answer 1

So you just need to integrate it.
\[\LARGE \int\limits 5x(x^2+1)^4 dx \]
using by parts.

Answer 2

erm

Answer 3

\[\Large \int\limits\limits 5x(x^2+1)^4 dx = 5x \int\limits (x^2+1)^4 - 5 \int\limits(x^2+1)^4dx\]

Answer 4

Alternatively,
\[\large (x^2+1)^4= 4C0 (x^2)4 + 4C1 (x^2)^3+4C2 (x^2)2 + 4C3 (x^2) + 4C4\]

Answer 5

wasteful
use substitution

Answer 6

oh yeh :P didn't strike :3 x^2+1=t woot

Answer 7

after factoring out 5
use u substitution and you are 2 steps away to completion

Answer 8

he is ryt

Answer 9

\[5 \int\limits x(x^2+1)^4dx \]
\[u = x^2+1;du = 2xdx\]
\[\frac{5}{2} \int\limits u^4du\]
do the rest

Answer 10

i still dont get it

Answer 11

This is a good time for advanced guessing.

You see the exponent 4 on the quantity \( (x^2 + 1) \). So ask yourself what would the derivative of \( (x^2 + 1)^5 \) look like? That's a chain rule problem.

Answer 12

The derivative of \((x^2 + 1)^5\) would be \( 5(x^2 +1)^4(2x) \) , right? That's just the chain rule. So in better order it would be \( 10x(x^2 +1)^4 \). Now compare that to your integrand.

Answer 13

Whoa! The derivative of \((x^2 + 1)^5\) is exactly twice your integrand. So what would the derivative of \({1 \over 2}(x^2 + 1)^5\) be?

Answer 14

here, read up on substitution method
https://www.math.hmc.edu/calculus/tutorials/substitution/

Answer 15

after completing the integration using the u-subtitution, you will need to return it back to what it was so you have an x and not U

Answer 16

there is no guess-work here, but just a use of the algebra you've learned your entire life and utilizing it more systematically

Answer 17

Advanced guessing, of course, is not guess-work. It is a way to avoid needless substitution.
The uniqueness theorem assures that if you come up with an antiderivative, it is the only answer, although it may not be in exactly the same form.

Answer 18

http://ocw.mit.edu/courses/mathematics/18-01sc-single-variable-calculus-fall-2010/unit-2-applications-of-differentiation/part-c-mean-value-theorem-antiderivatives-and-differential-equations/session-38-integration-by-substitution/

See clip 2 on this page, Prof. Jerison's "recommended method" = "advanced guessing"

Answer 19

if i had something like integral x*e^(x^2) ... can we solve it like this:

u=e^(x^2)

du=(2x*e^(x^2))

Answer 20

what should i do next?

Answer 21

just let u = x^2 , du =2xdx you have it becomes
\[\dfrac{1}{2}\int e^udu\]
It's easy now, right?

Answer 22

where did the 1/2 come from? @OOOPS

Answer 23

hey, du = 2x dx, --> xdx = 1/2 du,

Answer 24

Your original problem is
\[\int xe^{x^2}dx\]
right?

Answer 25

correct

Answer 26

u=e^(x^2)

du=x dx

Answer 27

If you let u = e^(x^2) , du = 2xe^(x^2) du
the whole integral is
\[\int \dfrac{1}{2}du =\dfrac{1}{2}u +C=\dfrac{1}{2}e^{x^2} +C\]

Answer 28

got it?