## ChrisV 3 years ago Find the indefinite integral. Please show work.

1. ChrisV

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

2. ChrisV

that (x+1)^2 is an exponent

3. nbouscal

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

4. ChrisV

yes

5. .Sam.

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

6. .Sam.

substitution u==x+1 du=dx $\text{}=\int\limits 5^{u^2} u \, du$ substitution again t=u^2 dt=2udu $\frac{1}{2}\int\limits 5^t \, dt$ $\frac{5^t}{2 \ln (5)}+c$ $\frac{5^{(x+1)^2}}{\log (25)}+c$

7. .Sam.

ln=log

8. nbouscal

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

9. nbouscal

Oh, Sam's way is a little easier.

10. ChrisV

well the answer the book gives me looks like

11. ChrisV

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

12. nbouscal

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

13. ChrisV

i just dont understand how they get to that

14. nbouscal

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)

15. ChrisV

i understand u substitution

16. ChrisV

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

17. nbouscal

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

18. nbouscal

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

19. ChrisV

ok here is where im confused

20. ChrisV

how does 5^t become 5^t/ln5

21. ChrisV

22. nbouscal

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

23. ChrisV

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

24. ChrisV

$1/2\int\limits5^t dt$

25. ChrisV

i would say that that is (ln5)t

26. ChrisV

and obviously im wrong lol

27. nbouscal

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

28. nbouscal

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

29. ChrisV

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

30. ChrisV

some things are just beyond me i guess

31. nbouscal

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

32. nbouscal

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

33. ChrisV

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

34. nbouscal

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

35. ChrisV

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