Find: ∫( (t^3+2t)/t )dt - Integration help?
I tried but i got an odd answer:
∫( (t^3+2t)/t )dt
∫( t^-1(t^3+2t) )dt
∫( t^-3 + 2t^-1 )dt
= t^-2/-2 + 2t^0/0 + c
But then you cant divide by 0 - otherwise the answer would technically be ∞?
More Questions From This User See All
Answers & Comments
Verified answer
Notice that (t^3+2t)/t=t^2+2
Which if you integrate you should get 2t+(t^3)/3+C
You are not integrating right at your last step
S = integral
Devide by t leaving
S[t^2+2)dt =t^3/3 +2t +k where K where k is a constant
1/3 t (6+t^2) + c
Im in cal 1 and just starting integrals, but that looks right to me, i literally just copied and pasted into wolfram alpha. I feel sorry for you if you're this far into calculus and dont know about it, it even shows you the steps, check it out:
http://www.wolframalpha.com/input/?i=+%E2%88%AB%28...
enable ln(t + 10) = u => t + 10 = e^u => dt = e^u du => necessary I = ? 2^u * e^u du using integration by using areas, I = e^u ? 2^u du - ? [d/du(e^u) ? 2^u du] du = a million/ln2 * 2^u e^u - a million/ln2 * ? 2^u * e^u du => (a million + a million/ln2) I = a million/ln2 * 2^u * e^u + c' => I = a million/(a million + ln2) * 2^u * e^u + c ... [c = c' * ln2/(a million + ln2)] => I = (t + 10) 2^[ln(t + 10)] / (a million + ln2) + c.
Y U NO ASK TEACHER?!