A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 4 years ago
Find the indefinite integral. Please show work.
anonymous
 4 years ago
Find the indefinite integral. Please show work.

This Question is Closed

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{?}^{?}(x+1)5^(x+1)^2\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0that (x+1)^2 is an exponent

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[ \int(x+1)5^{(x+1)^2}dx \]That?

.Sam.
 4 years ago
Best ResponseYou've already chosen the best response.1\[\Huge \int\limits_{}^{}(x+1)(5)^{(x+1)^2}\]

.Sam.
 4 years ago
Best ResponseYou've already chosen the best response.1substitution u==x+1 du=dx \[\text{}=\int\limits 5^{u^2} u \, du\] substitution again t=u^2 dt=2udu \[\frac{1}{2}\int\limits 5^t \, dt\] \[\frac{5^t}{2 \ln (5)}+c\] \[\frac{5^{(x+1)^2}}{\log (25)}+c\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Okay, so u=x+1 and we have \[ \int u\cdot 5^{u^2}du=\int u\cdot e^{u^2\log5}du=\frac{1}{2\log 5}\int 2\log5u\cdot e^{u^2\log5}du=\frac{e^{u^2\log5}}{2\log 5}=\frac{5^{(x+1)^2}}{2\log5} \]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Oh, Sam's way is a little easier.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0well the answer the book gives me looks like

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[1/2(5^{(x+1)^2}/\ln5)+C\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yep, that's the exact same thing as my answer and Sam's answer, just written slightly different.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i just dont understand how they get to that

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Which 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)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i understand u substitution

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.05^u^2 = (ln5)u^2 right?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[ 5^{u^2}=e^{(\log5)u^2} \]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[ \int a^xdx=\int e^{x\log a}dx=\frac{1}{\log a}\int(\log a) e^{x\log a}dx=\frac{e^{x\log a}}{\log a}=\frac{a^x}{\log a} \\\int a^xdx=\frac{a^x}{\log a} \]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok here is where im confused

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0how does 5^t become 5^t/ln5

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0He integrates it, using the formula I derived in my last comment.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0well thats where im lost. I do not know how to integrate that

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[1/2\int\limits5^t dt\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i would say that that is (ln5)t

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0and obviously im wrong lol

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I just showed you how to integrate that in my comment, did you not see that? For \(a\) as any constant.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\int a^xdx=\int e^{x\log a}dx=\frac{1}{\log a}\int(\log a) e^{x\log a}dx=\frac{e^{x\log a}}{\log a}=\frac{a^x}{\log a} \\\int a^xdx=\frac{a^x}{\log a}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0k ill write that down and try to wrap my brain around it

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0some things are just beyond me i guess

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0You just have to remember that \(e^{x^y}=e^{xy}\), which makes it so that \(a^x=(e^{\log a})^x=e^{x\log a}\)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Sorry, that first part should read \((e^x)^y=e^{xy}\)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0luckily i have a 99.8 average in this class so if i miss this on the final it shouldnt hurt much >.<

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Just revisit the rules for differentiating/integrating exponential and logarithmic functions. They come in handy for a lot of tricky integrals.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i try, i think my brain is overloaded. It is finals week
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.