How do you determine which is U and which is DV in solving integration by parts?

Question

How do you solve the integral of (e^x - x)^2 dx and the integral of xln2x dx ? How can you use uv - integral of vdu to solve these problems?

your help is greatly appreciated

Answer 1

Try this: This is the order in which you should pick 'u', the the other is 'dv'.

Log

Inverse trig

Algebraic

Trig

Exponential

Meaning that "u" should always be log first, then inverse trig, then algebraic terms, then trig, and last exponential.

For the first integral:

Let's foil : (e^x - x)(e^x - x)

∫ (e^2x - 2xe^x + x² ) dx

As you can see, we can split this up into 3 different integrals, making it less likely to make mistakes. Then, when we do integrate all three, we can simply add them back together.

  

∫ e^2x dx

here, let u = 2x, so that du = 2 dx

So all we need is to put a 2 beside the dx, and we'll have du. We must also put a 1/2 out front, so that we don't change the integral. Thus,

1/2 ∫ e^u du = 1/2 e^(2x)  <~~ This is one part.

For the second, let's do the easy integral:

∫ x²  dx  = 1/3 x³       <~~ This is the second part.

Finally, we have:

- ∫ 2x*e^x  dx

As you can see, we must use parts. Like I said, choose u as LIATE. Log > inverse trig > algebraic>trig>exponent

So let u = 2x         ....dv = -e^x dx

      .........du = 2dx     .... v = -e^x

Remembering the formula: ∫ udv = uv - ∫ vdu

= -2x*e^x + ∫ e^x*2dx

= -2xe^x + 2 ∫ e^x dx

= -2x*e^x + 2e^x

We can factor out a 2e^x, leaving the third piece of our first total integral to be:

2e^x (1-x) + C

Now, putthing the three pieces together,

∫ (e^x - x)² dx  = x³/3 + e^(2x)/2 + 2e^x(1-x) + C  <~~Answer

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For the second integral, use the LIATE again.

∫ x * ln(2x) dx

u = ln(2x)   .....dv = x

du = 1/x dx      v = x²/2

∫ udv = uv - ∫ vdu

= ln(2x) * x²/2 - ∫ x²/2 * 1/x dx

= x²/2 * ln(2x) - ∫ x/2 dx

= x²/2 * ln(2x) -  x² / 4 + C

Finally, we can factor out an: x²/2

x²/2 ( ln(2x) - 1/2) + C   <~~Answer

Hope this helps!