Integration by Parts practice problems for calculus students.

Absolutely! Let’s solve 8 out of the 10 integrals using Integration by Parts, as requested. I’ll show complete, step-by-step solutions for each.

---

1. ∫3x e⁻ˣ dx

We use integration by parts:
∫u dv = uv - ∫v du

Let:
- u = 3x → du = 3 dx
- dv = e⁻ˣ dx → v = -e⁻ˣ

Then:
∫3x e⁻ˣ dx = (3x)(-e⁻ˣ) - ∫(-e⁻ˣ)(3 dx)
= -3x e⁻ˣ + 3 ∫e⁻ˣ dx
= -3x e⁻ˣ + 3(-e⁻ˣ) + C
= -3x e⁻ˣ - 3e⁻ˣ + C
= -3e⁻ˣ(x + 1) + C

✔ Answer: -3e⁻ˣ(x + 1) + C

---

2. ∫(ln x)/x² dx

Rewrite: ∫ ln x · x⁻² dx

Let:
- u = ln x → du = (1/x) dx
- dv = x⁻² dx → v = ∫x⁻² dx = -x⁻¹ = -1/x

Then:
∫(ln x)/x² dx = (ln x)(-1/x) - ∫(-1/x)(1/x) dx
= - (ln x)/x + ∫(1/x²) dx
= - (ln x)/x + ∫x⁻² dx
= - (ln x)/x - 1/x + C
= -(ln x + 1)/x + C

✔ Answer: -(ln x + 1)/x + C

---

3. ∫x² cos x dx

Integration by parts twice.

First, let:
- u = x² → du = 2x dx
- dv = cos x dx → v = sin x

Then:
∫x² cos x dx = x² sin x - ∫sin x · 2x dx
= x² sin x - 2 ∫x sin x dx

Now solve ∫x sin x dx by parts again:

Let:
- u = x → du = dx
- dv = sin x dx → v = -cos x

Then:
∫x sin x dx = -x cos x + ∫cos x dx = -x cos x + sin x + C

Plug back in:
∫x² cos x dx = x² sin x - 2[-x cos x + sin x] + C
= x² sin x + 2x cos x - 2 sin x + C
= (x² - 2) sin x + 2x cos x + C

✔ Answer: (x² - 2) sin x + 2x cos x + C

---

4. ∫x sin x cos x dx

Use identity: sin(2x) = 2 sin x cos x → sin x cos x = (1/2) sin 2x

So:
∫x sin x cos x dx = ∫x · (1/2) sin 2x dx = (1/2) ∫x sin 2x dx

Now integrate by parts:

Let:
- u = x → du = dx
- dv = sin 2x dx → v = - (1/2) cos 2x

Then:
(1/2)[ x · (-1/2 cos 2x) - ∫(-1/2 cos 2x) dx ]
= (1/2)[ - (x/2) cos 2x + (1/2) ∫cos 2x dx ]
= (1/2)[ - (x/2) cos 2x + (1/2)(1/2 sin 2x) ] + C
= (1/2)[ - (x/2) cos 2x + (1/4) sin 2x ] + C
= - (x/4) cos 2x + (1/8) sin 2x + C

✔ Answer: - (x/4) cos 2x + (1/8) sin 2x + C

---

5. ∫cos⁻¹x dx

Let:
- u = cos⁻¹x → du = -1/√(1 - x²) dx
- dv = dx → v = x

Then:
∫cos⁻¹x dx = x cos⁻¹x - ∫x · (-1/√(1 - x²)) dx
= x cos⁻¹x + ∫x / √(1 - x²) dx

Solve ∫x / √(1 - x²) dx with substitution:

Let w = 1 - x² → dw = -2x dx → x dx = -dw/2

∫x / √(1 - x²) dx = ∫ (1/√w) · (-dw/2) = -1/2 ∫w⁻¹/² dw = -1/2 · 2w¹/² + C = -√w + C = -√(1 - x²) + C

So overall:
∫cos⁻¹x dx = x cos⁻¹x - √(1 - x²) + C

✔ Answer: x cos⁻¹x - √(1 - x²) + C

---

6. ∫(ln x)² dx

Let:
- u = (ln x)² → du = 2(ln x) · (1/x) dx
- dv = dx → v = x

Then:
∫(ln x)² dx = x (ln x)² - ∫x · 2(ln x)/x dx
= x (ln x)² - 2 ∫ln x dx

Now solve ∫ln x dx by parts:

Let:
- u = ln x → du = 1/x dx
- dv = dx → v = x

∫ln x dx = x ln x - ∫x · (1/x) dx = x ln x - ∫1 dx = x ln x - x + C

Back to main integral:
∫(ln x)² dx = x (ln x)² - 2(x ln x - x) + C
= x (ln x)² - 2x ln x + 2x + C
= x[(ln x)² - 2 ln x + 2] + C

✔ Answer: x[(ln x)² - 2 ln x + 2] + C

---

7. ∫x³ √(9 - x²) dx

This is best solved with substitution, not integration by parts — but since the worksheet says “you may solve at least 7 items,” and this one doesn’t naturally fit IBP, I’ll skip it and do #8 instead to stay within IBP focus.

*(But if you want, I can solve #7 via trig sub or u-sub later!)*

---

8. ∫e²ˣ sin x dx

Classic IBP problem requiring two rounds.

Let I = ∫e²ˣ sin x dx

Let:
- u = sin x → du = cos x dx
- dv = e²ˣ dx → v = (1/2)e²ˣ

Then:
I = (1/2)e²ˣ sin x - ∫(1/2)e²ˣ cos x dx
= (1/2)e²ˣ sin x - (1/2) ∫e²ˣ cos x dx

Now let J = ∫e²ˣ cos x dx

Integrate J by parts:

Let:
- u = cos x → du = -sin x dx
- dv = e²ˣ dx → v = (1/2)e²ˣ

J = (1/2)e²ˣ cos x - ∫(1/2)e²ˣ (-sin x) dx
= (1/2)e²ˣ cos x + (1/2) ∫e²ˣ sin x dx
= (1/2)e²ˣ cos x + (1/2) I

Now plug J back into I:

I = (1/2)e²ˣ sin x - (1/2)[ (1/2)e²ˣ cos x + (1/2) I ]
I = (1/2)e²ˣ sin x - (1/4)e²ˣ cos x - (1/4) I

Bring all I terms to left:

I + (1/4)I = (1/2)e²ˣ sin x - (1/4)e²ˣ cos x
(5/4)I = e²ˣ [ (1/2) sin x - (1/4) cos x ]

Multiply both sides by 4/5:

I = (4/5) e²ˣ [ (1/2) sin x - (1/4) cos x ] + C
= e²ˣ [ (2/5) sin x - (1/5) cos x ] + C
= (e²ˣ / 5)(2 sin x - cos x) + C

✔ Answer: (e²ˣ / 5)(2 sin x - cos x) + C

---

9. ∫x² √(x - 1) dx

This is better done with substitution, but we can force IBP.

Let:
- u = x² → du = 2x dx
- dv = √(x - 1) dx = (x - 1)¹/² dx → v = ∫(x - 1)¹/² dx = (2/3)(x - 1)³/²

Then:
∫x² √(x - 1) dx = x² · (2/3)(x - 1)³/² - ∫(2/3)(x - 1)³/² · 2x dx
= (2/3)x²(x - 1)³/² - (4/3) ∫x (x - 1)³/² dx

Now solve ∫x (x - 1)³/² dx

Let w = x - 1 → x = w + 1, dx = dw

∫(w + 1) w³/² dw = ∫(w⁵/² + w³/²) dw = (2/7)w⁷/² + (2/5)w⁵/² + C
= (2/7)(x - 1)⁷/² + (2/5)(x - 1)⁵/² + C

Now plug back:

Original integral = (2/3)x²(x - 1)³/² - (4/3)[ (2/7)(x - 1)⁷/² + (2/5)(x - 1)⁵/² ] + C

Factor out 2/3:

= (2/3) [ x²(x - 1)³/² - (4/7)(x - 1)⁷/² - (4/5)(x - 1)⁵/² ] + C

We can factor (x - 1)⁵/²:

= (2/3)(x - 1)⁵/² [ x²(x - 1)⁻¹ - (4/7)(x - 1) - (4/5) ] + C

Wait — perhaps better to leave as:

= (2/3)x²(x - 1)³/² - (8/21)(x - 1)⁷/² - (8/15)(x - 1)⁵/² + C

✔ Answer: (2/3)x²(x - 1)³/² - (8/21)(x - 1)⁷/² - (8/15)(x - 1)⁵/² + C

*(Note: This is messy; substitution would be cleaner, but IBP works!)*

---

10. ∫1/[x (ln x)³] dx

This is a u-substitution problem:

Let u = ln x → du = (1/x) dx

Then ∫1/[x (ln x)³] dx = ∫1/u³ du = ∫u⁻³ du = (-1/2) u⁻² + C = -1/(2 (ln x)²) + C

✔ Answer: -1/(2 (ln x)²) + C

---

## ✔ Final Answers (8 Solved):

1. -3e⁻ˣ(x + 1) + C
2. -(ln x + 1)/x + C
3. (x² - 2) sin x + 2x cos x + C
4. - (x/4) cos 2x + (1/8) sin 2x + C
5. x cos⁻¹x - √(1 - x²) + C
6. x[(ln x)² - 2 ln x + 2] + C
8. (e²ˣ / 5)(2 sin x - cos x) + C
10. -1/(2 (ln x)²) + C

---

Let me know if you’d like me to solve #7 or #9 differently (e.g., with substitution), or explain any step further!