Integral of this simple function

[iScience]

im trying to find...

∫te^(5t)dt

i did integration by parts..

u=e^5t
dv=tdt

∫te^(5t)dt={e^(5t)}(t)-5∫(t){e^(5t)}dt

collected the integral terms on one side to get..

6∫te^(5t)dt={e^(5t)}(t)

∫te^(5t)dt= {e^(5t)}(t)/6

but wolfram alpha says otherwise. they don't give out the steps freely anymore. please help

 

[Ackbach]

Use ILATE for integration by parts:

Inverse trig
Logarithmic
Algebraic
Trig
Exponential

Whichever function is higher on the list gets to be u, and dv is whichever function is lower. You have a t, which is algebraic, and you have an exp(5t), which is exponential. Try that.

 

[Ray Vickson]

im trying to find...

∫te^(5t)dt

i did integration by parts..

u=e^5t
dv=tdt

∫te^(5t)dt={e^(5t)}(t)-5∫(t){e^(5t)}dt

collected the integral terms on one side to get..

6∫te^(5t)dt={e^(5t)}(t)

∫te^(5t)dt= {e^(5t)}(t)/6

but wolfram alpha says otherwise. they don't give out the steps freely anymore. please help

With u = e^(5t) and dv = t dt, you will get

∫ t e^(5t) dt = (1/2)t^2 e^(5t) - (5/2) ∫ t^2 e^(5t) dt,

so you have just made it worse! You have the wrong u and dv.

 

[vanhees71]

Define the function
f(t,x)=\int \mathrm{d} t \exp(t x)=\frac{1}{x} \exp(t x).
Then you have
\partial_x f(t,x)=\int \mathrm{d} t \exp(t x)=-\frac{1}{x^2} \exp(t x)+\frac{t}{x} \exp(t x).
Setting x=5 you find
\int \mathrm{d} t \exp(5t)=\exp(5t) \left [\frac{t}{5}-\frac{1}{25} \right].

 

[Millennial]

Define the function
f(t,x)=\int \mathrm{d} t \exp(t x)=\frac{1}{x} \exp(t x).
Then you have
\partial_x f(t,x)=\int \mathrm{d} t \exp(t x)=-\frac{1}{x^2} \exp(t x)+\frac{t}{x} \exp(t x).
Setting x=5 you find
\int \mathrm{d} t \exp(5t)=\exp(5t) \left [\frac{t}{5}-\frac{1}{25} \right].

There is a bit of faulty notation here. After partially differentiating f with respect to t, you should have an additional t factor inside the integral, which is what I think you wanted, as your solution seems to be using that. I believe you have forgot to type it..?

 

[CAF123]

Inverse trig
Logarithmic
Algebraic
Trig
Exponential

Whichever function is higher on the list gets to be u, and dv is whichever function is lower..

I never memorised this list - I just look at what could possibly be u and dv and then decide what would ultimately simplify the integral. As Ray said, by choosing dv = t dt, all you do is introduce greater powers of t whereas choosing u = t, when you differentiate it, you simply get 1 and hence is more likely to simply the integral. I think (i may be wrong) this is how that ILATE list was created - after some experience, it was put together as an easy way to decide on u and dv which would lead to a simpler integral.

 

[vanhees71]

There is a bit of faulty notation here. After partially differentiating f with respect to t, you should have an additional t factor inside the integral, which is what I think you wanted, as your solution seems to be using that. I believe you have forgot to type it..?

Sure, that's a typo :-(. Correctly it should read
\partial_x f(t,x)=\int \mathrm{d} t \; t \exp(t x)=\ldots