Kkutie7
  • Kkutie7
How do you integrate:
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Kkutie7
  • Kkutie7
\[\int\limits\frac{t}{\sqrt{t+1}}\]
tkhunny
  • tkhunny
Have you considered a substitution?
Kkutie7
  • Kkutie7
well u substitution doesn't really work and I can't remember any trig ones so I'm kinda stuck

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

Kkutie7
  • Kkutie7
wait I might be able to do u substitution...
tkhunny
  • tkhunny
ALWAYS try whatever is under the radical. It might just work. u = t + 1 Go!
Kkutie7
  • Kkutie7
wouldn't that make \[\int \frac{t}{\sqrt{u}}du\rightarrow \int \frac{u-1}{u}du\]
Kkutie7
  • Kkutie7
oops that was supposed to be a square on the second one
Kkutie7
  • Kkutie7
what about integration by parts better?
tkhunny
  • tkhunny
?? You're almost done. Why are you seeking another method?
tkhunny
  • tkhunny
You can try u^2 = (t+1)^2 if you like. We already know that t + 1 > 0, so that's okay. Quiz: How do we know that t + 1 > 0?
Kkutie7
  • Kkutie7
well the only way I can think of continuing is like this: \[\int \frac{u-1}{\sqrt{u}}du\rightarrow \int u^{-\frac{1}{2}}(u-1)du\]
anonymous
  • anonymous
Now distribute
tkhunny
  • tkhunny
What's wrong with that? \(u^{1/2} - u^{-1/2}\). That should be easy.
Kkutie7
  • Kkutie7
ok \[\int u^{\frac{1}{2}}-u^{\frac{-1}{2}}du\rightarrow\frac{2}{3}u^{\frac{3}{2}}-2u^{\frac{1}{2}du}+C\]
Kkutie7
  • Kkutie7
* du shouldn't be there
anonymous
  • anonymous
looks good
anonymous
  • anonymous
Oh one more thing, substitute back for t
anonymous
  • anonymous
\[\frac{2}{3}(t+1)^{\frac{3}{2}}-2\sqrt{t+1}+C\]
Kkutie7
  • Kkutie7
the answer is supposed to be \[\frac{2}{3}(t-2)\sqrt{t+1}+C\]
Kkutie7
  • Kkutie7
I really don't get it
Astrophysics
  • Astrophysics
\[\int\limits \frac{ t }{ (t+1)^{1/2} }dt \]\[u = t+1 \implies du = dt, ~~~~ t = u-1\]\[\int\limits \frac{ (u-1) }{ u^{1/2} } du \implies \int\limits \frac{ u }{ u^{1/2} }-\frac{ 1 }{ u^{1/2} } du\] \[\implies \int\limits u^{1/2}-u^{-1/2} du\] \[\frac{ u^{1/2+1} }{ 1/2+1 } - \frac{ u^{-1/2+1} }{ -1/2+1 } +C\] \[\implies \frac{ 2}{ 3 } u^{3/2}-2u^{1/2}+C\] plug in our original u subs \[\frac{ 2 }{ 3 }(t+1)^{3/2}-2(t+1)^{1/2}+C\]
Astrophysics
  • Astrophysics
It's the same thing
rvc
  • rvc
yep \[\rm u~get~:\\2/3u^{(3/2)}-2u^{1/2}+C\\~=2u^{1/2}(u/3-1)+C~\\~=2(\sqrt{t+1})(\frac{t+1-3}{3})+C~\\~=2(\sqrt{t+1})(\frac{t-2}{3})+C\]
tkhunny
  • tkhunny
LaTeX Note: If you use "\left(\right)" pairs, rather than "()" pairs, you get better parentheses that expand with what they contain. ( \ d f r a c { t - 2 } { 3 } ) ==> \((\dfrac{t-2}{3})\) \ l e f t ( \ d f r a c { t - 2 } { 3 } \ r I g h t ) ==> \(\left(\dfrac{t-2}{3}\right)\)

Looking for something else?

Not the answer you are looking for? Search for more explanations.