Question

Ok I am not sure if this was a mistake but I found it on a practice exam I was studying and it looks really hard to do:

Integral of (25x(5x+1)^30)

Expanding everything out would take way too long and I see no way to integrate that via u-substitution.

Answer 1

i would try integrating by parts , with u=25x and dv=(5x+1)^30

Answer 2

ok thanks, this is a calculus I course so idk why it was on the test then since i thought integration by parts was a calc II topic

Answer 3

you havent covered integration by parts?

Answer 4

I learned it back in high school, but integration by parts is not a topic that should be in calc I

Answer 5

um then i dunno how else would u solve this other than integrating by parts

Answer 6

the problem looks crazy if you input it in wolframalpha

Answer 7

they have it expanded there thats why

Answer 8

what i get for solution is \[\frac{ 5x }{ 31 }(5x+1)^{31}-\frac{ (5x+1)^{32} }{ 992 }\]

Answer 9

\[25x(5x+1)^{30}\]
\[25x\sum_{n=0}^{30}\binom{30}{n}(5x)^{30-n}(1)^{n}\]
\[5^2x^1\sum_{n=0}^{30}\binom{30}{n}(5)^{30-n}~(x)^{30-n}\]
\[\sum_{n=0}^{30}\binom{30}{n}(5)^{32-n}~(x)^{31-n}\]
integrating part by part .... we get
\[\sum_{n=0}^{30}\binom{30}{n}(5)^{32-n}~\frac{(x)^{31-n+1}}{31-n+1}\]
seems to be something like that is my idea

Answer 10

Figured it out. Set u=5x+1 and integrate it via u-substiturion. To solve the x probelm set x= u-1/5

Answer 11

well yeah, if you want to be uncreative ... i guess thats a route you can take :)