Interactive Integration by Parts Worksheet featuring 12 calculus problems to solve using integration by parts.

Let's solve each of these integrals step by step using integration techniques, especially focusing on integration by parts where appropriate. We'll go through each one carefully.

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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 $

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

Now integrate $ \int e^x \cos x\, dx $ by parts:

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

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

Now substitute back:

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

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

Bring the integral to the left:

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

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

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

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

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} $

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

$$
= \frac{3}{4}x^{4/3} \ln x - \frac{3}{4} \cdot \frac{x^{4/3}}{4/3} = \frac{3}{4}x^{4/3} \ln x - \frac{3}{4} \cdot \frac{3}{4}x^{4/3}
= \frac{3}{4}x^{4/3} \ln x - \frac{9}{16}x^{4/3}
$$

Now multiply by $ \frac{1}{\ln 4} $:

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

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

Integration by parts (use tabular method or repeated):

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

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

Now integrate $ \int x^2 \cos(4x)\, dx $:

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

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

Now $ \int x \sin(4x)\, dx $:

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

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

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 first expression:

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

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

$$
\boxed{\int x^3 \sin(4x)\, dx = -\frac{1}{4}x^3 \cos(4x) + \frac{3}{16}x^2 \sin(4x) + \frac{3}{32}x \cos(4x) - \frac{3}{128}\sin(4x) + C}
$$

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

Similar to #1.

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

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

Now $ \int e^x \cos(2x)\, dx $:

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

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

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 to left:

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

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

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

$ \log_2 x = \frac{\ln x}{\ln 2} $, so:

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

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

$$
= \frac{1}{\ln 2} \left( \frac{x^4}{4} \ln x - \int \frac{x^4}{4} \cdot \frac{1}{x} dx \right)
= \frac{1}{\ln 2} \left( \frac{x^4}{4} \ln x - \frac{1}{4} \int x^3 dx \right)
= \frac{1}{\ln 2} \left( \frac{x^4}{4} \ln x - \frac{1}{4} \cdot \frac{x^4}{4} \right)
$$

$$
= \frac{1}{\ln 2} \left( \frac{x^4}{4} \ln x - \frac{x^4}{16} \right)
= \frac{x^4}{4\ln 2} \ln x - \frac{x^4}{16\ln 2} + C
$$

$$
\boxed{\int x^3 \log_2 x\, dx = \frac{x^4}{4\ln 2} \ln x - \frac{x^4}{16\ln 2} + C}
$$

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

Let $ u = \ln(3x) = \ln 3 + \ln x $. Then $ du = \frac{1}{x} dx $

But we need $ dx $, so let’s try substitution:

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

But $ dx = x\, du $, and $ x = e^u / 3 $? Better:

Since $ u = \ln(3x) $, then $ 3x = e^u $ → $ x = \frac{e^u}{3} $ → $ dx = \frac{e^u}{3} du $

So:

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

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

So:

$$
= \frac{3x}{6} (\cos(\ln(3x)) + \sin(\ln(3x))) = \frac{x}{2} (\cos(\ln(3x)) + \sin(\ln(3x)))
$$

$$
\boxed{\int \cos(\ln 3x)\, dx = \frac{x}{2} \left( \cos(\ln 3x) + \sin(\ln 3x) \right) + C}
$$

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

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

But we have $ x^3 $ in numerator, not $ x^2 $. Let’s write $ x^3 = u - 1 $

Then $ dx = \frac{du}{3x^2} $, but we still have $ x^2 $ in denominator.

Try expressing $ x^3 = u - 1 $, and $ x^2 = ? $ — not helpful.

Alternative idea: Let $ u = x^3 + 1 $, then $ du = 3x^2 dx $

But we have $ x^3 dx $, not $ x^2 dx $. So express $ x^3 dx $ as $ x \cdot x^2 dx $

Note: $ x^3 dx = x \cdot x^2 dx $, but not helpful.

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

Then $ x^2 dx = \frac{1}{3} du $

But we need $ x^3 dx $. Note that $ x^3 = u - 1 $, and $ x^3 dx = (u - 1) \cdot \frac{dx}{x^{-3}} $? Not good.

Wait: $ x^3 dx = x \cdot (x^2 dx) = x \cdot \frac{1}{3} du $

But $ x = (u - 1)^{1/3} $, so:

$$
x^3 dx = x \cdot x^2 dx = (u - 1)^{1/3} \cdot \frac{1}{3} du
$$

So:

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

This is messy. Perhaps better to use substitution: $ u = \sqrt{x^3 + 1} $, so $ u^2 = x^3 + 1 $

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

Still not directly helpful.

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

But we want $ x^3 dx $. Let’s write:

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

Then $ x^3 = u - 1 $

We need $ x^3 dx = (u - 1) dx $

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

So $ dx = \frac{du}{3(u - 1)^{2/3}} $

Thus:

$$
x^3 dx = (u - 1) \cdot \frac{du}{3(u - 1)^{2/3}} = \frac{1}{3} (u - 1)^{1 - 2/3} du = \frac{1}{3} (u - 1)^{1/3} du
$$

Now the integral becomes:

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

Still complicated. This may require hypergeometric functions. But likely intended to be solved via trig sub or rationalizing.

Alternatively, let’s try a different approach.

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

But we can’t easily express $ x^3 dx $ in terms of $ du $.

Wait: Try integrating by parts?

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

Not promising.

Alternatively, notice that derivative of $ x^3 + 1 $ is $ 3x^2 $, so maybe split:

Let’s try $ w = x^3 + 1 $, $ dw = 3x^2 dx $

But our numerator is $ x^3 dx $. Write $ x^3 = w - 1 $, and $ dx = \frac{dw}{3x^2} $

Then $ x^3 dx = (w - 1) \cdot \frac{dw}{3x^2} = \frac{w - 1}{3x^2} dw $

But $ x^2 = (x^3)^{2/3} = (w - 1)^{2/3} $, so:

$$
x^3 dx = \frac{w - 1}{3(w - 1)^{2/3}} dw = \frac{1}{3}(w - 1)^{1/3} dw
$$

So:

$$
\int \frac{x^3}{\sqrt{x^3 + 1}} dx = \int \frac{1}{\sqrt{w}} \cdot \frac{1}{3}(w - 1)^{1/3} dw = \frac{1}{3} \int w^{-1/2} (w - 1)^{1/3} dw
$$

This is a standard form for beta function or hypergeometric, but not elementary.

Wait — perhaps it was meant to be $ \int \frac{x^2}{\sqrt{x^3 + 1}} dx $? That would be easy.

But as written, this might be not elementary.

But let’s double-check: Is there a substitution?

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

But we don't have $ x^2 dx $. But suppose we write:

Let’s consider $ \int \frac{x^3}{\sqrt{x^3 + 1}} dx = \int \frac{(x^3 + 1) - 1}{\sqrt{x^3 + 1}} dx = \int \sqrt{x^3 + 1} dx - \int \frac{1}{\sqrt{x^3 + 1}} dx $

Now both integrals are non-elementary (elliptic integrals). So likely a typo.

Alternatively, if it were $ \int \frac{x^2}{\sqrt{x^3 + 1}} dx $, then let $ u = x^3 + 1 $, $ du = 3x^2 dx $, so:

$$
= \frac{1}{3} \int u^{-1/2} du = \frac{1}{3} \cdot 2u^{1/2} = \frac{2}{3} \sqrt{x^3 + 1} + C
$$

But as written, this integral does not have an elementary antiderivative.

So unless there's a typo, we may skip or note it's non-elementary.

But let’s assume it’s a typo and it's $ x^2 $ instead of $ x^3 $. Or perhaps proceed numerically.

But since it's on a worksheet, likely a mistake.

Alternatively, maybe $ \int \frac{x^2}{\sqrt{x^3 + 1}} dx $ — that's what I’ll assume.

But as written: $ \int \frac{x^3}{\sqrt{x^3 + 1}} dx $ — no elementary antiderivative.

So possibly not solvable with elementary methods.

But let’s move on.

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

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

Then:

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

Now $ u = x - 1 $, so:

$$
= \frac{2}{5}(x - 1)^{5/2} + \frac{2}{3}(x - 1)^{3/2} + C
$$

$$
\boxed{\int x \sqrt{x - 1}\, dx = \frac{2}{5}(x - 1)^{5/2} + \frac{2}{3}(x - 1)^{3/2} + C}
$$

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

Integration by parts:

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

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

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

So:

$$
\boxed{\int \tan^{-1} x\, dx = x \tan^{-1} x - \frac{1}{2} \ln(1 + x^2) + C}
$$

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

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

Then $ 2x = 2(u + 5) $

So:

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

Now $ u = x - 5 $:

$$
\boxed{\int \frac{2x}{(x - 5)^3} dx = -\frac{2}{x - 5} - \frac{5}{(x - 5)^2} + C}
$$

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

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

So:

$$
= \frac{1}{\ln 6} \int \ln x\, dx = \frac{1}{\ln 6} (x \ln x - x) + C
$$

$$
\boxed{\int \log_6 x\, dx = \frac{x}{\ln 6} (\ln x - 1) + C}
$$

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

This is not straightforward.

Let $ u = x^2 $ → $ du = 2x dx $, but we have $ x^2 dx $, not $ x dx $

So $ x^2 dx = x \cdot x dx = x \cdot \frac{du}{2} $, but $ x = \sqrt{u} $, so:

$$
x^2 dx = \sqrt{u} \cdot \frac{du}{2} = \frac{1}{2} u^{1/2} du
$$

So:

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

This is not elementary — involves Fresnel integrals or special functions.

So no elementary antiderivative.

But perhaps the problem meant $ \int x \cos(x^2) dx $? That would be:

Let $ u = x^2 $, $ du = 2x dx $ → $ \frac{1}{2} \int \cos u\, du = \frac{1}{2} \sin u + C = \frac{1}{2} \sin(x^2) + C $

But as written, $ x^2 \cos(x^2) $, no elementary antiderivative.

So likely not solvable in elementary terms.

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Summary of Answers:

1. $ \boxed{\frac{e^x}{2}(\sin x - \cos x) + C} $

2. $ \boxed{\frac{1}{\ln 4} \left( \frac{3}{4}x^{4/3} \ln x - \frac{9}{16}x^{4/3} \right) + C} $

3. $ \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. $ \boxed{\frac{e^x}{5}(\sin(2x) - 2\cos(2x)) + C} $

5. $ \boxed{\frac{x^4}{4\ln 2} \ln x - \frac{x^4}{16\ln 2} + C} $

6. $ \boxed{\frac{x}{2} \left( \cos(\ln 3x) + \sin(\ln 3x) \right) + C} $

7. $ \boxed{\text{No elementary antiderivative}} $ (likely typo)

8. $ \boxed{\frac{2}{5}(x - 1)^{5/2} + \frac{2}{3}(x - 1)^{3/2} + C} $

9. $ \boxed{x \tan^{-1} x - \frac{1}{2} \ln(1 + x^2) + C} $

10. $ \boxed{-\frac{2}{x - 5} - \frac{5}{(x - 5)^2} + C} $

11. $ \boxed{\frac{x}{\ln 6} (\ln x - 1) + C} $

12. $ \boxed{\text{No elementary antiderivative}} $

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Let me know if you'd like clarification on any of them!