Interactive Integration by Parts Worksheet featuring 12 calculus problems for practice.

It looks like you've uploaded a worksheet titled "Interactive Integration by Parts Worksheet" with 12 integrals to solve. Since the image contains only the list of problems and no specific instruction to solve just one, I'll go ahead and solve each problem step-by-step, using appropriate techniques such as integration by parts, substitution, or other standard methods.

Let’s solve all 12 integrals:

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1. $ \int e^x \sin x \, dx $

Use integration by parts twice.

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

Then:
$$
\int e^x \sin x\,dx = e^x \sin x - \int e^x \cos x\,dx
$$

Now integrate $ \int e^x \cos x\,dx $ by parts:
- $ u = \cos x $ → $ du = -\sin x\,dx $
- $ dv = e^x dx $ → $ v = e^x $

So:
$$
\int e^x \cos x\,dx = e^x \cos x + \int e^x \sin x\,dx
$$

Substitute back:
$$
\int e^x \sin x\,dx = e^x \sin x - \left( e^x \cos x + \int e^x \sin x\,dx \right)
$$
$$
\int e^x \sin x\,dx = e^x \sin x - e^x \cos x - \int e^x \sin x\,dx
$$

Add $ \int e^x \sin x\,dx $ to both sides:
$$
2\int e^x \sin x\,dx = e^x (\sin x - \cos x)
$$
$$
\int e^x \sin x\,dx = \frac{e^x}{2}(\sin x - \cos x) + C
$$

✔ Answer: $ \boxed{\frac{e^x}{2}(\sin x - \cos x) + C} $

---

2. $ \int \sqrt[3]{x} \log_4 x \, dx $

Note: $ \sqrt[3]{x} = x^{1/3} $, and $ \log_4 x = \frac{\ln x}{\ln 4} $

So:
$$
\int x^{1/3} \cdot \frac{\ln x}{\ln 4} dx = \frac{1}{\ln 4} \int x^{1/3} \ln x\,dx
$$

Use integration by parts:
- $ u = \ln x $ → $ du = \frac{1}{x} dx $
- $ dv = x^{1/3} dx $ → $ v = \frac{x^{4/3}}{4/3} = \frac{3}{4}x^{4/3} $

Then:
$$
\int x^{1/3} \ln x\,dx = \frac{3}{4}x^{4/3} \ln x - \int \frac{3}{4}x^{4/3} \cdot \frac{1}{x} dx
= \frac{3}{4}x^{4/3} \ln x - \frac{3}{4} \int x^{1/3} dx
$$

Now:
$$
\int x^{1/3} dx = \frac{x^{4/3}}{4/3} = \frac{3}{4}x^{4/3}
$$

So:
$$
\int x^{1/3} \ln x\,dx = \frac{3}{4}x^{4/3} \ln x - \frac{3}{4} \cdot \frac{3}{4}x^{4/3} = \frac{3}{4}x^{4/3} \ln x - \frac{9}{16}x^{4/3}
$$

Multiply by $ \frac{1}{\ln 4} $:
$$
\int \sqrt[3]{x} \log_4 x\,dx = \frac{1}{\ln 4} \left( \frac{3}{4}x^{4/3} \ln x - \frac{9}{16}x^{4/3} \right) + C
$$

Factor:
$$
= \frac{3}{4\ln 4} x^{4/3} \left( \ln x - \frac{3}{4} \right) + C
$$

✔ Answer: $ \boxed{\frac{3}{4\ln 4} x^{4/3} \left( \ln x - \frac{3}{4} \right) + C} $

---

3. $ \int x^3 \sin(4x)\,dx $

Use integration by parts, repeated.

Let:
- $ u = x^3 $ → $ du = 3x^2 dx $
- $ dv = \sin(4x) dx $ → $ v = -\frac{1}{4}\cos(4x) $

Then:
$$
\int x^3 \sin(4x)\,dx = -\frac{1}{4}x^3 \cos(4x) + \frac{3}{4} \int x^2 \cos(4x)\,dx
$$

Now integrate $ \int x^2 \cos(4x)\,dx $:
- $ u = x^2 $ → $ du = 2x dx $
- $ dv = \cos(4x) dx $ → $ v = \frac{1}{4}\sin(4x) $

So:
$$
\int x^2 \cos(4x)\,dx = \frac{1}{4}x^2 \sin(4x) - \frac{1}{2} \int x \sin(4x)\,dx
$$

Now $ \int x \sin(4x)\,dx $:
- $ u = x $ → $ du = dx $
- $ dv = \sin(4x) dx $ → $ v = -\frac{1}{4}\cos(4x) $

$$
\int x \sin(4x)\,dx = -\frac{1}{4}x \cos(4x) + \frac{1}{4} \int \cos(4x)\,dx = -\frac{1}{4}x \cos(4x) + \frac{1}{16} \sin(4x)
$$

Now back-substitute:
$$
\int x^2 \cos(4x)\,dx = \frac{1}{4}x^2 \sin(4x) - \frac{1}{2} \left( -\frac{1}{4}x \cos(4x) + \frac{1}{16} \sin(4x) \right)
= \frac{1}{4}x^2 \sin(4x) + \frac{1}{8}x \cos(4x) - \frac{1}{32} \sin(4x)
$$

Now plug into original:
$$
\int x^3 \sin(4x)\,dx = -\frac{1}{4}x^3 \cos(4x) + \frac{3}{4} \left( \frac{1}{4}x^2 \sin(4x) + \frac{1}{8}x \cos(4x) - \frac{1}{32} \sin(4x) \right)
$$

Simplify:
$$
= -\frac{1}{4}x^3 \cos(4x) + \frac{3}{16}x^2 \sin(4x) + \frac{3}{32}x \cos(4x) - \frac{3}{128} \sin(4x) + C
$$

✔ Answer: $ \boxed{-\frac{1}{4}x^3 \cos(4x) + \frac{3}{16}x^2 \sin(4x) + \frac{3}{32}x \cos(4x) - \frac{3}{128} \sin(4x) + C} $

---

4. $ \int e^x \sin(2x)\,dx $

Similar to #1, use integration by parts twice.

Let:
- $ u = \sin(2x) $ → $ du = 2\cos(2x) dx $
- $ dv = e^x dx $ → $ v = e^x $

So:
$$
\int e^x \sin(2x)\,dx = e^x \sin(2x) - 2 \int e^x \cos(2x)\,dx
$$

Now integrate $ \int e^x \cos(2x)\,dx $:
- $ u = \cos(2x) $ → $ du = -2\sin(2x) dx $
- $ dv = e^x dx $ → $ v = e^x $

$$
\int e^x \cos(2x)\,dx = e^x \cos(2x) + 2 \int e^x \sin(2x)\,dx
$$

Substitute back:
$$
\int e^x \sin(2x)\,dx = e^x \sin(2x) - 2 \left( e^x \cos(2x) + 2 \int e^x \sin(2x)\,dx \right)
= e^x \sin(2x) - 2e^x \cos(2x) - 4 \int e^x \sin(2x)\,dx
$$

Bring integral to left:
$$
5 \int e^x \sin(2x)\,dx = e^x (\sin(2x) - 2\cos(2x))
$$

$$
\int e^x \sin(2x)\,dx = \frac{e^x}{5} (\sin(2x) - 2\cos(2x)) + C
$$

✔ Answer: $ \boxed{\frac{e^x}{5} (\sin(2x) - 2\cos(2x)) + C} $

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5. $ \int x^3 \log_2 x\,dx $

Recall: $ \log_2 x = \frac{\ln x}{\ln 2} $

So:
$$
\int x^3 \log_2 x\,dx = \frac{1}{\ln 2} \int x^3 \ln x\,dx
$$

Use integration by parts:
- $ u = \ln x $ → $ du = \frac{1}{x} dx $
- $ dv = x^3 dx $ → $ v = \frac{x^4}{4} $

Then:
$$
\int x^3 \ln x\,dx = \frac{x^4}{4} \ln x - \int \frac{x^4}{4} \cdot \frac{1}{x} dx = \frac{x^4}{4} \ln x - \frac{1}{4} \int x^3 dx
= \frac{x^4}{4} \ln x - \frac{1}{4} \cdot \frac{x^4}{4} = \frac{x^4}{4} \ln x - \frac{x^4}{16}
$$

Now multiply by $ \frac{1}{\ln 2} $:
$$
\int x^3 \log_2 x\,dx = \frac{1}{\ln 2} \left( \frac{x^4}{4} \ln x - \frac{x^4}{16} \right) + C
= \frac{x^4}{4\ln 2} \ln x - \frac{x^4}{16\ln 2} + C
$$

✔ Answer: $ \boxed{\frac{x^4}{4\ln 2} \ln x - \frac{x^4}{16\ln 2} + C} $

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6. $ \int \cos(\ln 3x)\,dx $

Let $ u = \ln(3x) $. Then:
- $ du = \frac{1}{x} dx $ → $ dx = x\,du $
- But $ x = \frac{e^u}{3} $? Better: $ u = \ln 3x = \ln 3 + \ln x $, so $ du = \frac{1}{x} dx $

But we need to express $ dx $ in terms of $ du $, but also $ x $ is involved.

Alternatively, let $ u = \ln(3x) $ → $ du = \frac{1}{x} dx $

So $ dx = x\,du $, but $ x = \frac{e^u}{3} $? Let's do substitution properly.

Let $ u = \ln(3x) $ → $ du = \frac{1}{x} dx $ → $ dx = x\,du $

But $ x = \frac{e^u}{3} $, since $ e^u = 3x $ → $ x = \frac{e^u}{3} $

So:
$$
\int \cos(u) \cdot dx = \int \cos(u) \cdot x\,du = \int \cos(u) \cdot \frac{e^u}{3} du = \frac{1}{3} \int e^u \cos u\,du
$$

This is similar to #1!

We know:
$$
\int e^u \cos u\,du = \frac{e^u}{2}(\cos u + \sin u) + C
$$

So:
$$
\frac{1}{3} \cdot \frac{e^u}{2}(\cos u + \sin u) = \frac{e^u}{6}(\cos u + \sin u)
$$

Now recall $ e^u = 3x $, $ u = \ln(3x) $

So $ \cos u = \cos(\ln(3x)) $, $ \sin u = \sin(\ln(3x)) $

Thus:
$$
\int \cos(\ln 3x)\,dx = \frac{3x}{6} (\cos(\ln 3x) + \sin(\ln 3x)) + C = \frac{x}{2} (\cos(\ln 3x) + \sin(\ln 3x)) + C
$$

✔ Answer: $ \boxed{\frac{x}{2} \left( \cos(\ln 3x) + \sin(\ln 3x) \right) + C} $

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7. $ \int \frac{x^3}{\sqrt{x^3 + 1}}\,dx $

Let $ u = x^3 + 1 $ → $ du = 3x^2 dx $

But numerator is $ x^3 $, not $ x^2 $. Try expressing $ x^3 = u - 1 $

But $ dx $ has $ x^2 $, so we need $ x^2 dx $

We have $ du = 3x^2 dx $ → $ x^2 dx = \frac{1}{3} du $

But we have $ x^3 dx $, which is $ x \cdot x^2 dx = x \cdot \frac{1}{3} du $

But $ x = (u - 1)^{1/3} $, messy.

Alternative idea: write $ x^3 = (x^3 + 1) - 1 = u - 1 $

So:
$$
\int \frac{x^3}{\sqrt{x^3 + 1}} dx = \int \frac{u - 1}{\sqrt{u}} \cdot \frac{dx}{?}
$$

But $ dx $ is not directly related.

Wait: $ u = x^3 + 1 $, $ du = 3x^2 dx $

But we want $ x^3 dx $. Not easy.

Try another approach: let $ u = x^3 + 1 $, then $ du = 3x^2 dx $

But $ x^3 dx = ? $

Note: $ x^3 dx = x \cdot x^2 dx $, and $ x^2 dx = \frac{1}{3} du $, so $ x^3 dx = x \cdot \frac{1}{3} du $

But $ x = (u - 1)^{1/3} $, so:
$$
\int \frac{x^3}{\sqrt{u}} dx = \int \frac{(u - 1)}{\sqrt{u}} \cdot \frac{x}{3} du = \frac{1}{3} \int \frac{u - 1}{\sqrt{u}} (u - 1)^{1/3} du
= \frac{1}{3} \int (u - 1)^{1 + 1/3} u^{-1/2} du
= \frac{1}{3} \int (u - 1)^{4/3} u^{-1/2} du
$$

Too complicated. Try a different substitution.

Let $ u = \sqrt{x^3 + 1} $ → $ u^2 = x^3 + 1 $

Differentiate: $ 2u\,du = 3x^2 dx $

Still not helpful.

Better: try substitution $ u = x^3 + 1 $, $ du = 3x^2 dx $

We need $ x^3 dx $, so write $ x^3 dx = x \cdot x^2 dx = x \cdot \frac{1}{3} du $

But $ x = (u - 1)^{1/3} $, so:
$$
\int \frac{x^3}{\sqrt{x^3 + 1}} dx = \int \frac{u - 1}{\sqrt{u}} \cdot \frac{x}{3} du = \frac{1}{3} \int \frac{u - 1}{\sqrt{u}} (u - 1)^{1/3} du = \frac{1}{3} \int (u - 1)^{4/3} u^{-1/2} du
$$

Still messy. Perhaps this is better done via trigonometric substitution or rationalizing.

Alternate idea: let $ u = x^3 + 1 $, $ du = 3x^2 dx $

But $ x^3 = u - 1 $, and $ dx = \frac{du}{3x^2} $

So:
$$
\int \frac{x^3}{\sqrt{u}} \cdot \frac{du}{3x^2} = \int \frac{u - 1}{\sqrt{u}} \cdot \frac{1}{3x^2} du
$$

But $ x^2 = (x^3)^{2/3} = (u - 1)^{2/3} $, so:
$$
= \frac{1}{3} \int \frac{u - 1}{\sqrt{u}} \cdot (u - 1)^{-2/3} du = \frac{1}{3} \int (u - 1)^{1 - 2/3} u^{-1/2} du = \frac{1}{3} \int (u - 1)^{1/3} u^{-1/2} du
$$

This is still hard. Maybe it's best to use substitution $ u = x^3 + 1 $, and accept that it becomes:
$$
\int \frac{x^3}{\sqrt{x^3 + 1}} dx = \int \frac{u - 1}{\sqrt{u}} \cdot \frac{dx}{?}
$$

Wait — perhaps let $ u = x^3 $?

Then $ du = 3x^2 dx $, again not helping.

Maybe this is meant for integration by parts?

Let me try:
Let $ u = x^3 $, $ dv = \frac{1}{\sqrt{x^3 + 1}} dx $

But $ dv $ is hard to integrate.

Alternatively, try substitution: $ u = x^3 + 1 $, $ du = 3x^2 dx $

But we have $ x^3 dx $, so write:
$$
x^3 dx = x \cdot x^2 dx = x \cdot \frac{1}{3} du
$$

And $ x = (u - 1)^{1/3} $, so:
$$
\int \frac{x^3}{\sqrt{u}} dx = \int \frac{u - 1}{\sqrt{u}} \cdot \frac{x}{3} du = \frac{1}{3} \int \frac{u - 1}{\sqrt{u}} (u - 1)^{1/3} du = \frac{1}{3} \int (u - 1)^{4/3} u^{-1/2} du
$$

Let $ w = u - 1 $, $ u = w + 1 $, $ dw = du $

Then:
$$
\frac{1}{3} \int w^{4/3} (w + 1)^{-1/2} dw
$$

This is not elementary. So likely this problem has a typo or requires advanced methods.

Wait — maybe try rationalizing or write as:

Let $ u = \sqrt{x^3 + 1} $, $ u^2 = x^3 + 1 $

Then $ 2u\,du = 3x^2 dx $

But again, not matching.

Perhaps the intended method is substitution $ u = x^3 + 1 $, and express $ x^3 = u - 1 $, but we can't avoid $ x $.

Alternatively, notice that:
$$
\frac{x^3}{\sqrt{x^3 + 1}} = \frac{(x^3 + 1) - 1}{\sqrt{x^3 + 1}} = \sqrt{x^3 + 1} - \frac{1}{\sqrt{x^3 + 1}}
$$

Ah! This is key!

So:
$$
\int \frac{x^3}{\sqrt{x^3 + 1}} dx = \int \left( \sqrt{x^3 + 1} - \frac{1}{\sqrt{x^3 + 1}} \right) dx
= \int (x^3 + 1)^{1/2} dx - \int (x^3 + 1)^{-1/2} dx
$$

These are non-elementary integrals. They cannot be expressed in terms of elementary functions.

So this integral does not have an elementary antiderivative.

✔ Answer: $ \boxed{\text{Non-elementary; no closed-form in elementary functions}} $

---

8. $ \int x \sqrt{x - 1} \, dx $

Let $ u = x - 1 $ → $ x = u + 1 $, $ dx = du $

Then:
$$
\int (u + 1) \sqrt{u} \, du = \int (u + 1) u^{1/2} du = \int (u^{3/2} + u^{1/2}) du
= \frac{u^{5/2}}{5/2} + \frac{u^{3/2}}{3/2} + C = \frac{2}{5} u^{5/2} + \frac{2}{3} u^{3/2} + C
$$

Now substitute back $ u = x - 1 $:
$$
= \frac{2}{5} (x - 1)^{5/2} + \frac{2}{3} (x - 1)^{3/2} + C
$$

✔ Answer: $ \boxed{\frac{2}{5} (x - 1)^{5/2} + \frac{2}{3} (x - 1)^{3/2} + C} $

---

9. $ \int \tan^{-1} x \, dx $

Integration by parts:
- $ u = \tan^{-1} x $ → $ du = \frac{1}{1 + x^2} dx $
- $ dv = dx $ → $ v = x $

So:
$$
\int \tan^{-1} x\,dx = x \tan^{-1} x - \int \frac{x}{1 + x^2} dx
$$

Now $ \int \frac{x}{1 + x^2} dx = \frac{1}{2} \ln(1 + x^2) + C $

So:
$$
= x \tan^{-1} x - \frac{1}{2} \ln(1 + x^2) + C
$$

✔ Answer: $ \boxed{x \tan^{-1} x - \frac{1}{2} \ln(1 + x^2) + C} $

---

10. $ \int \frac{2x}{(x - 5)^3} dx $

Let $ u = x - 5 $ → $ x = u + 5 $, $ dx = du $

Then:
$$
\int \frac{2(u + 5)}{u^3} du = 2 \int \left( \frac{u}{u^3} + \frac{5}{u^3} \right) du = 2 \int \left( u^{-2} + 5u^{-3} \right) du
= 2 \left( \frac{u^{-1}}{-1} + 5 \cdot \frac{u^{-2}}{-2} \right) + C
= 2 \left( -u^{-1} - \frac{5}{2} u^{-2} \right) + C
= -\frac{2}{u} - \frac{5}{u^2} + C
$$

Now substitute back $ u = x - 5 $:
$$
= -\frac{2}{x - 5} - \frac{5}{(x - 5)^2} + C
$$

✔ Answer: $ \boxed{-\frac{2}{x - 5} - \frac{5}{(x - 5)^2} + C} $

---

11. $ \int \log_6 x\,dx $

$ \log_6 x = \frac{\ln x}{\ln 6} $

So:
$$
\int \log_6 x\,dx = \frac{1}{\ln 6} \int \ln x\,dx
$$

We know $ \int \ln x\,dx = x \ln x - x + C $

So:
$$
= \frac{1}{\ln 6} (x \ln x - x) + C = \frac{x}{\ln 6} (\ln x - 1) + C
$$

✔ Answer: $ \boxed{\frac{x}{\ln 6} (\ln x - 1) + C} $

---

12. $ \int x^2 \cos(x^2)\,dx $

This is tricky. Note: $ \cos(x^2) $ is not easily integrable.

Let $ u = x^2 $ → $ du = 2x dx $

But we have $ x^2 \cos(x^2) dx $, which is $ u \cos u \cdot \frac{dx}{?} $

We need $ x dx $, but we have $ x^2 dx $

Note: $ x^2 dx = x \cdot x dx = x \cdot \frac{1}{2} du $

But $ x = \sqrt{u} $, so:
$$
\int x^2 \cos(x^2) dx = \int u \cos u \cdot \frac{1}{2} \cdot \frac{1}{\sqrt{u}} du = \frac{1}{2} \int \sqrt{u} \cos u\,du
$$

Which is $ \frac{1}{2} \int u^{1/2} \cos u\,du $ — this is not elementary.

This integral involves the Fresnel integral and is not expressible in elementary functions.

✔ Answer: $ \boxed{\text{Non-elementary; no closed-form in elementary functions}} $

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✔ Final Summary of Answers:

| Problem | Answer |
|--------|--------|
| 1 | $ \frac{e^x}{2}(\sin x - \cos x) + C $ |
| 2 | $ \frac{3}{4\ln 4} x^{4/3} \left( \ln x - \frac{3}{4} \right) + C $ |
| 3 | $ -\frac{1}{4}x^3 \cos(4x) + \frac{3}{16}x^2 \sin(4x) + \frac{3}{32}x \cos(4x) - \frac{3}{128} \sin(4x) + C $ |
| 4 | $ \frac{e^x}{5} (\sin(2x) - 2\cos(2x)) + C $ |
| 5 | $ \frac{x^4}{4\ln 2} \ln x - \frac{x^4}{16\ln 2} + C $ |
| 6 | $ \frac{x}{2} \left( \cos(\ln 3x) + \sin(\ln 3x) \right) + C $ |
| 7 | Non-elementary |
| 8 | $ \frac{2}{5} (x - 1)^{5/2} + \frac{2}{3} (x - 1)^{3/2} + C $ |
| 9 | $ x \tan^{-1} x - \frac{1}{2} \ln(1 + x^2) + C $ |
| 10 | $ -\frac{2}{x - 5} - \frac{5}{(x - 5)^2} + C $ |
| 11 | $ \frac{x}{\ln 6} (\ln x - 1) + C $ |
| 12 | Non-elementary |

Let me know if you'd like detailed steps for any specific problem!