MHB Complex Integrals (for me at least)

[filiphenrique]

Hey!

How do I integrate ∫tln√(t+1) and ∫4te^(2-0,3t)?
Thanks in advance.

 

[DreamWeaver]

On that first integral, you might try using $$\log x^a = a\log x$$, and re-write it in the equivalent form$$\int t\log\sqrt{1+t}\, dt = \frac{1}{2}\, \int t\log(1+t)\, dt$$

Next, try the substitution $$y=1+t?$$NOTE: I always use "log" instead of "ln", so where you wrote "ln" I've written "log".

 

[DreamWeaver]

For the second integral, I'm not entirely sure what you mean. Is it

$$4\, \int e^{2-0.3 t} \, dt = ?$$

If so, then it might be worth separating the exponential term into two parts; one that contains the variable "t", and must - therefore - be kept inside the integral sign, while the other is a constant $$(e^{x+y}=e^xe^y)$$:

$$\int e^{2-0.3 t}\, dt = e^2\, \int e^{-0.3 t}\, dt$$

 

[soroban]

Hello, filiphenrique!

Integrate these "by parts".
I'll do the first one.

I \;=\;\int t \ln\sqrt{t+1}\,dt

\begin{Bmatrix}u &=& \tfrac{1}{2}\ln(t+1) && dv &=& t\,dt \\ du &=& \tfrac{dt}{2(t+1)} && v &=& \tfrac{1}{2}t^2\end{Bmatrix}

I \;=\;\tfrac{1}{4}t^2\ln(t+1) - \tfrac{1}{4}\int \frac{t^2}{t+1}\,dt

I \;=\;\tfrac{1}{4}t^2\ln(t+1) - \tfrac{1}{4}\int\left(t - 1 + \frac{1}{t+1}\right)\,dt
I \;=\;\tfrac{1}{4}t^2\ln(t+1) - \tfrac{1}{4}\left[\tfrac{1}{2}t^2 - t + \ln(t+1)\right] + C

I \;=\;\tfrac{1}{4}t^2\ln(t+1) - \tfrac{1}{8}t^2 + \tfrac{1}{4}t - \tfrac{1}{4}\ln(t+1) + C

 

[filiphenrique]

For the second integral, I'm not entirely sure what you mean. Is it

$$4\, \int e^{2-0.3 t} \, dt = ?$$

If so, then it might be worth separating the exponential term into two parts; one that contains the variable "t", and must - therefore - be kept inside the integral sign, while the other is a constant $$(e^{x+y}=e^xe^y)$$:

$$\int e^{2-0.3 t}\, dt = e^2\, \int e^{-0.3 t}\, dt$$

Hey, thanks for the reply. I got what you meant in the first integral I posted. But in the second integral, We really have the t variable before the e^(2-0,3t). Based on the way you start solving the integral, can I put the 4 and the e^2 off and integrate te^0,3t by parts? Thanks!

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By the way, how do I write integrals the way you guys do here?

 

[DreamWeaver]

Hiya! :D

Yes, you certainly can integrate that last one by parts, starting with $$(d/dt)e^{at} = ?$$

To show integrals the way I posted above, write

View attachment 2579

to display as

$$ \int f(x)\, dx = g(x) $$
And as for the integral itself, if $$a$$ and $$b$$ are constants, then

$$\int e^{a+bx}\,dx = \int (e^a\, e^{bx})\, dx = e^a\int e^{bx}\, dx$$

Can you solve it from there? :D