Question

Find the indefinite integral. Please show work.

Answer 1

\[\int\limits_{?}^{?}(x+1)5^(x+1)^2\]

Answer 2

that (x+1)^2 is an exponent

Answer 3

\[
\int(x+1)5^{(x+1)^2}dx
\]That?

Answer 4

yes

Answer 5

\[\Huge \int\limits_{}^{}(x+1)(5)^{(x+1)^2}\]

Answer 6

substitution
u==x+1 du=dx
\[\text{}=\int\limits 5^{u^2} u \, du\]
substitution again
t=u^2 dt=2udu

\[\frac{1}{2}\int\limits 5^t \, dt\]
\[\frac{5^t}{2 \ln (5)}+c\]
\[\frac{5^{(x+1)^2}}{\log (25)}+c\]

Answer 7

ln=log

Answer 8

Okay, so u=x+1 and we have \[
\int u\cdot 5^{u^2}du=\int u\cdot e^{u^2\log5}du=\frac{1}{2\log 5}\int 2\log5u\cdot e^{u^2\log5}du=\frac{e^{u^2\log5}}{2\log 5}=\frac{5^{(x+1)^2}}{2\log5}
\]

Answer 9

Oh, Sam's way is a little easier.

Answer 10

well the answer the book gives me looks like

Answer 11

\[1/2(5^{(x+1)^2}/\ln5)+C\]

Answer 12

Yep, that's the exact same thing as my answer and Sam's answer, just written slightly different.

Answer 13

i just dont understand how they get to that

Answer 14

Which part of the method Sam and I used did you not follow? (Sam's is a bit easier to follow because he does substitution twice and I only used it once)

Answer 15

i understand u substitution

Answer 16

5^u^2 = (ln5)u^2 right?

Answer 17

\[
5^{u^2}=e^{(\log5)u^2}
\]

Answer 18

\[
\int a^xdx=\int e^{x\log a}dx=\frac{1}{\log a}\int(\log a) e^{x\log a}dx=\frac{e^{x\log a}}{\log a}=\frac{a^x}{\log a} \\\int a^xdx=\frac{a^x}{\log a}
\]

Answer 19

ok here is where im confused

Answer 20

how does 5^t become 5^t/ln5

Answer 21

in sams answer

Answer 22

He integrates it, using the formula I derived in my last comment.

Answer 23

well thats where im lost. I do not know how to integrate that

Answer 24

\[1/2\int\limits5^t dt\]

Answer 25

i would say that that is (ln5)t

Answer 26

and obviously im wrong lol

Answer 27

I just showed you how to integrate that in my comment, did you not see that? For \(a\) as any constant.

Answer 28

\[\int a^xdx=\int e^{x\log a}dx=\frac{1}{\log a}\int(\log a) e^{x\log a}dx=\frac{e^{x\log a}}{\log a}=\frac{a^x}{\log a} \\\int a^xdx=\frac{a^x}{\log a}\]

Answer 29

k ill write that down and try to wrap my brain around it

Answer 30

some things are just beyond me i guess

Answer 31

You just have to remember that \(e^{x^y}=e^{xy}\), which makes it so that \(a^x=(e^{\log a})^x=e^{x\log a}\)

Answer 32

Sorry, that first part should read \((e^x)^y=e^{xy}\)

Answer 33

luckily i have a 99.8 average in this class so if i miss this on the final it shouldnt hurt much >.<

Answer 34

Just revisit the rules for differentiating/integrating exponential and logarithmic functions. They come in handy for a lot of tricky integrals.

Answer 35

i try, i think my brain is overloaded. It is finals week