Question

So this one is almost exactly like the last one but I'm still doing something wrong

Answer 1

\[\int\limits_{}^{}x^4\sqrt{5+x^5}dx\]
u=\[5+x^{5}\]
\[\frac{ du }{ dx }=5x ^{4}\]
du=5x^4 dx

Answer 2

yes
\[
I=\int\sqrt{5+x^5}(x^4dx)=\int\sqrt{u}\frac{du}{5}
\]

Answer 3

\[\frac{ 1 }{ 5 }du=x^{4}dx\]
\[\frac{ 1 }{ 5 } \int\limits_{}^{}\sqrt{u} du\]

Answer 4

bingo

Answer 5

so why isn't webworks accepting it?

Answer 6

show your final answer

Answer 7

(sqrt(5+x^5))/5+C

Answer 8

see

Answer 9

\[
I={1\over5}\int u^{1\over2}du={1\over5}\frac{u^{3\over2}}{3\over2}+C\\
I={1\over5}{2\over3}u\sqrt{u}+C\\
I={2\over15}(1+x^5)\sqrt{1+x^5}+C
\]

Answer 10

now I'm lost again

Answer 11

the first statement is the one right after the substitution, right?

Answer 12

i think you did everything but skipped the actual "integration": part

Answer 13

so I get that you changed the \[\sqrt{u} \to u^{1/2}\]
and then took the antiderivative of that

Answer 14

ye

Answer 15

which is \[\frac{ 1 }{ 5 }*\frac{ 2 }{ 3 }u ^{3/2}+C\]

Answer 16

yep

Answer 17

but then why/how did you go back to \[\sqrt{u}\]

Answer 18

ooh \[u^{3/2}=u\times u^{1/2}=u\sqrt{u}\]

Answer 19

I do not like exponent.. this way looks waaay cooler

Answer 20

ok so then keeping it my way would be
\[\frac{ 1 }{ 5 }\frac{ 2 }{ 3 }(5+x^5)^{2/3}+C\]

Answer 21

yes .
perfectly fine

Answer 22

which is sloppy, but I get it and webworks accepts it

Answer 23

@heradog in your last post you have the exponent as 2/3 not 3/2.

Answer 24

oops, I had it right everywhere else, thanks