Questions: Integration by Parts

Question 1

To evaluate ∫ x·eˣ dx by integration by parts, which choice of u and dv is most effective?

• A. u = eˣ, dv = x dx
• B. u = x, dv = eˣ dx
• C. u = xeˣ, dv = dx
• D. Neither; u-substitution is needed instead

Answer: (B)

Choosing u = x (Algebraic, per LIATE) and dv = eˣ dx gives du = dx and v = eˣ, so the formula yields x·eˣ − ∫eˣ dx = xeˣ − eˣ + C. The reverse choice (u = eˣ) differentiates eˣ into eˣ again, making the new integral ∫x·eˣ dx — the same integral you started with, which is no progress.

Question 2

Evaluating ∫ ln(x) dx by integration by parts requires identifying a second function being multiplied by ln(x).

• T. True
• F. False

Answer: False

You can write ∫ ln(x) dx as ∫ ln(x)·1 dx and set u = ln(x), dv = 1 dx. Then du = (1/x) dx and v = x, giving x·ln(x) − ∫x·(1/x) dx = x·ln(x) − x + C. The factor of 1 is a perfectly valid dv, and this move — taking u to be the entire original integrand — is a standard technique for logarithms and inverse trig functions.

Question 3

When integrating ∫ eˣ·sin(x) dx by parts twice, you obtain an equation of the form I = (expression) − I. How do you use this to find the integral?

Model answer: Treat I = ∫ eˣ sin(x) dx as an unknown. After applying IBP twice you get I = eˣ(sin x − cos x) − I. Adding I to both sides gives 2I = eˣ(sin x − cos x), so I = eˣ(sin x − cos x)/2 + C.

The cyclical case arises because differentiating sin(x) twice brings back sin(x), and similarly for eˣ. Rather than being stuck in an infinite loop, the equation I = f(x) − I is algebraically solvable. This is a feature, not a failure — it gives a clean closed-form answer.