Integration by Parts worksheet with six problems and substitution hints.

We are given 6 indefinite integrals to solve using Integration by Parts, with the substitutions for `u` and `dv` already provided. The formula for integration by parts is:

> ∫ u dv = u v - ∫ v du

Let’s solve each one step-by-step.

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1) ∫ e^x sin(x) dx, with u = sin(x), dv = e^x dx

- Then:
du = cos(x) dx
v = ∫ e^x dx = e^x

Apply integration by parts:

∫ e^x sin(x) dx = sin(x) e^x - ∫ e^x cos(x) dx

Now we need to integrate ∫ e^x cos(x) dx — again use integration by parts.

Let:
u = cos(x), dv = e^x dx
→ du = -sin(x) dx, v = e^x

So:

∫ e^x cos(x) dx = cos(x) e^x - ∫ e^x (-sin(x)) dx
= e^x cos(x) + ∫ e^x sin(x) dx

Now plug back into the first equation:

∫ e^x sin(x) dx = e^x sin(x) - [e^x cos(x) + ∫ e^x sin(x) dx]

Bring the integral from the right to the left:

∫ e^x sin(x) dx + ∫ e^x sin(x) dx = e^x sin(x) - e^x cos(x)

2 ∫ e^x sin(x) dx = e^x (sin(x) - cos(x))

Divide both sides by 2:

Answer:
> ∫ e^x sin(x) dx = (e^x / 2)(sin(x) - cos(x)) + C

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2) ∫ x² e^(4x) dx, with u = x², dv = e^(4x) dx

- du = 2x dx
- v = ∫ e^(4x) dx = (1/4) e^(4x)

Apply integration by parts:

∫ x² e^(4x) dx = x² * (1/4)e^(4x) - ∫ (1/4)e^(4x) * 2x dx
= (x² e^(4x))/4 - (1/2) ∫ x e^(4x) dx

Now compute ∫ x e^(4x) dx — again use integration by parts.

Let:
u = x, dv = e^(4x) dx
→ du = dx, v = (1/4)e^(4x)

So:

∫ x e^(4x) dx = x*(1/4)e^(4x) - ∫ (1/4)e^(4x) dx
= (x e^(4x))/4 - (1/4)*(1/4)e^(4x)
= (x e^(4x))/4 - (1/16)e^(4x)

Plug back in:

∫ x² e^(4x) dx = (x² e^(4x))/4 - (1/2)[(x e^(4x))/4 - (1/16)e^(4x)]
= (x² e^(4x))/4 - (x e^(4x))/8 + (1/32)e^(4x) + C

Factor out e^(4x)/32:

= (e^(4x)/32)(8x² - 4x + 1) + C

Answer:
> ∫ x² e^(4x) dx = (e^(4x)/32)(8x² - 4x + 1) + C

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3) ∫ ln(x)/x dx, with u = ln(x), dv = (1/x) dx

Wait — this is a trick! Let's check:

If u = ln(x), then du = (1/x) dx
And dv = (1/x) dx → v = ∫ (1/x) dx = ln|x| (but that’s messy and leads to circularity)

Actually, this integral is much simpler with substitution!

Let’s do it correctly — even though the worksheet says to use parts, let’s see what happens if we follow the instructions.

Given: u = ln(x), dv = (1/x) dx

Then:
du = (1/x) dx
v = ∫ (1/x) dx = ln|x| (we’ll drop absolute value for x > 0)

So:

∫ ln(x)/x dx = u v - ∫ v du = ln(x) * ln(x) - ∫ ln(x) * (1/x) dx
= [ln(x)]² - ∫ ln(x)/x dx

Bring the integral to the left:

∫ ln(x)/x dx + ∫ ln(x)/x dx = [ln(x)]²
→ 2 ∫ ln(x)/x dx = [ln(x)]²

Thus:

Answer:
> ∫ ln(x)/x dx = (1/2)[ln(x)]² + C

✔ This works, but note: you could have done it directly with substitution w = ln(x), dw = (1/x)dx → ∫ w dw = (1/2)w² + C.

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4) ∫ x² sin(10x) dx, with u = x², dv = sin(10x) dx

- du = 2x dx
- v = ∫ sin(10x) dx = (-1/10) cos(10x)

Apply integration by parts:

∫ x² sin(10x) dx = x² * (-1/10) cos(10x) - ∫ (-1/10) cos(10x) * 2x dx
= - (x² cos(10x))/10 + (1/5) ∫ x cos(10x) dx

Now compute ∫ x cos(10x) dx — again use integration by parts.

Let:
u = x, dv = cos(10x) dx
→ du = dx, v = (1/10) sin(10x)

So:

∫ x cos(10x) dx = x*(1/10) sin(10x) - ∫ (1/10) sin(10x) dx
= (x sin(10x))/10 + (1/100) cos(10x) + C

Now plug back in:

∫ x² sin(10x) dx = - (x² cos(10x))/10 + (1/5)[ (x sin(10x))/10 + (1/100) cos(10x) ] + C
= - (x² cos(10x))/10 + (x sin(10x))/50 + (1/500) cos(10x) + C

Combine like terms:

= - (x² cos(10x))/10 + (x sin(10x))/50 + cos(10x)/500 + C

Factor cos(10x):

= cos(10x) [ -x²/10 + 1/500 ] + (x sin(10x))/50 + C

Or leave as is.

Answer:
> ∫ x² sin(10x) dx = - (x² cos(10x))/10 + (x sin(10x))/50 + (cos(10x))/500 + C

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5) ∫ x⁵ √(x³ + 9) dx, with u = x³, dv = x² √(x³ + 9) dx

Wait — this is strange. Usually in integration by parts, we pick u and dv so that du and v are easier. Here, dv = x² √(x³ + 9) dx — which is actually easy to integrate via substitution!

Let’s follow the instruction.

Given: u = x³ → du = 3x² dx
dv = x² √(x³ + 9) dx

Find v = ∫ x² √(x³ + 9) dx

Use substitution: let w = x³ + 9 → dw = 3x² dx → x² dx = dw/3

So:

v = ∫ √w * (dw/3) = (1/3) ∫ w^(1/2) dw = (1/3)(2/3) w^(3/2) = (2/9)(x³ + 9)^(3/2)

Now apply integration by parts:

∫ u dv = u v - ∫ v du
= x³ * (2/9)(x³ + 9)^(3/2) - ∫ (2/9)(x³ + 9)^(3/2) * 3x² dx
= (2/9) x³ (x³ + 9)^(3/2) - (2/3) ∫ x² (x³ + 9)^(3/2) dx

Now compute the new integral: ∫ x² (x³ + 9)^(3/2) dx

Again, substitute: w = x³ + 9 → dw = 3x² dx → x² dx = dw/3

So:

∫ x² (x³ + 9)^(3/2) dx = ∫ w^(3/2) * (dw/3) = (1/3) * (2/5) w^(5/2) = (2/15)(x³ + 9)^(5/2)

Now plug back:

Original integral = (2/9) x³ (x³ + 9)^(3/2) - (2/3) * (2/15)(x³ + 9)^(5/2) + C
= (2/9) x³ (x³ + 9)^(3/2) - (4/45)(x³ + 9)^(5/2) + C

Factor out (2/45)(x³ + 9)^(3/2):

= (2/45)(x³ + 9)^(3/2) [ 5x³ - 2(x³ + 9) ] + C
= (2/45)(x³ + 9)^(3/2) [5x³ - 2x³ - 18]
= (2/45)(x³ + 9)^(3/2)(3x³ - 18)
= (2/45)(x³ + 9)^(3/2) * 3(x³ - 6)
= (6/45)(x³ + 9)^(3/2)(x³ - 6)
= (2/15)(x³ + 9)^(3/2)(x³ - 6) + C

Answer:
> ∫ x⁵ √(x³ + 9) dx = (2/15)(x³ + 9)^(3/2)(x³ - 6) + C

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6) ∫ x √(x + 2) dx, with u = x, dv = √(x + 2) dx

- du = dx
- v = ∫ √(x + 2) dx = ∫ (x + 2)^(1/2) dx = (2/3)(x + 2)^(3/2)

Apply integration by parts:

∫ x √(x + 2) dx = x * (2/3)(x + 2)^(3/2) - ∫ (2/3)(x + 2)^(3/2) dx

Compute the remaining integral:

∫ (x + 2)^(3/2) dx = (2/5)(x + 2)^(5/2)

So:

= (2/3)x (x + 2)^(3/2) - (2/3) * (2/5)(x + 2)^(5/2) + C
= (2/3)x (x + 2)^(3/2) - (4/15)(x + 2)^(5/2) + C

Factor out (2/15)(x + 2)^(3/2):

= (2/15)(x + 2)^(3/2) [5x - 2(x + 2)] + C
= (2/15)(x + 2)^(3/2)(5x - 2x - 4)
= (2/15)(x + 2)^(3/2)(3x - 4) + C

Answer:
> ∫ x √(x + 2) dx = (2/15)(x + 2)^(3/2)(3x - 4) + C

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## ✔ Final Answers:

1) ∫ e^x sin(x) dx = (e^x / 2)(sin(x) - cos(x)) + C

2) ∫ x² e^(4x) dx = (e^(4x)/32)(8x² - 4x + 1) + C

3) ∫ ln(x)/x dx = (1/2)[ln(x)]² + C

4) ∫ x² sin(10x) dx = - (x² cos(10x))/10 + (x sin(10x))/50 + (cos(10x))/500 + C

5) ∫ x⁵ √(x³ + 9) dx = (2/15)(x³ + 9)^(3/2)(x³ - 6) + C

6) ∫ x √(x + 2) dx = (2/15)(x + 2)^(3/2)(3x - 4) + C

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These solutions follow the provided substitutions and demonstrate careful application of integration by parts, sometimes requiring multiple applications or algebraic simplification.