# Evaluate integration

CellCoree
Evaluate $$\int t^{14} ln(t) dt$$

using integration by parts and formula $$uv-\intv du$$

u=ln(t)
du=1/t
dv=t^14
v=t^15/15

lnt*t^15/15-$$\int$$t^15/15*1/t

lnt*t^15/15-t^14*log(t)

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Gold Member
First step is fine. You're over-thinking the second step:

$$\int{\frac{t^{15}}{15} \frac{dt}{t}} = \int{\frac{t^{14}}{15}dt = \frac{t^{15}}{225}$$

$$\int{\left(\ln t\right) t^{14} dt} = \frac{t^{15}}{15} \left(\ln t - \frac{1}{15} \right)$$

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Homework Helper
Gold Member
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CellCoree said:
Evaluate $$\int t^{14} ln(t) dt$$

using integration by parts and formula $$uv-\intv du$$

u=ln(t)
du=1/t
dv=t^14
v=t^15/15

lnt*t^15/15-$$\int$$t^15/15*1/t

lnt*t^15/15-t^14*log(t)

DO NOT USE THIS HORRIBLE MIXTURE OF INTEGRATION BY PARTS AND INTEGRATION BY SUBSTITUTION!
Write:
$$\int{uv'}dt=uv-\int{u'v}dt$$
$$u=ln(t)\to{u}'=\frac{1}{t}$$
$$v'=t^{14}\to{v}=\frac{1}{15}t^{15}$$

CellCoree
arildno said:
DO NOT USE THIS HORRIBLE MIXTURE OF INTEGRATION BY PARTS AND INTEGRATION BY SUBSTITUTION!
Write:
$$\int{uv'}dt=uv-\int{u'v}dt$$
$$u=ln(t)\to{u}'=\frac{1}{t}$$
$$v'=t^{14}\to{v}=\frac{1}{15}t^{15}$$

what do you mean? your setup is just like mines

Homework Helper
Gold Member
Dearly Missed
CellCoree said:
what do you mean? your setup is just like mines
Nope!
1. "Using formula uv-du"
Complete nonsense
If you want to use this method properly, write:
udv=d(uv)-vdu
2.Furthermore
you use a meaningless constuction like du=1/t
If you want to use this method, you must write du=1/t*dt
3. Your integral lacks dt, and is therefore meaningless as well

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CellCoree
jamesrc said:
First step is fine. You're over-thinking the second step:

$$\int{\frac{t^{15}}{15} \frac{dt}{t}} = \int{\frac{t^{14}}{15}dt = \frac{t^{15}}{225}$$

$$\int{\left(\ln t\right) t^{14} dt} = \frac{t^{15}}{15} \left(\ln t - \frac{1}{15} \right)$$

ok i know how you got $$\int{\frac{t^{15}}{15} \frac{dt}{t}} = \int{\frac{t^{14}}{15}dt$$

but how did you get

$$\int{\frac{t^{14}}{15}dt = \frac{t^{15}}{225}$$

i know you got the anti-derv. of it, and i tried to look in my book for a form like that, but i can't find it. can you explain to me how you did that?

for $$\int{\frac{t^{14}}{15}dt$$ i used u-du sub.

u=14
du=u+1

so... t^u/du but i can't seem to find that form in the book

$$\int \frac{t^{14}}{15} dt = \frac{1}{15}\int t^{14} dt = \frac{1}{15}\frac{t^{15}}{15} = \frac{t^{15}}{(15)^{2}} = \frac{t^{15}}{225}$$