Collection of integral calculus problems requiring integration techniques.

To solve the given integrals, we will use various techniques such as integration by parts, substitution, and trigonometric identities. Let's go through each integral step by step.

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(1) \( \int x e^x \, dx \)

Solution:
We use integration by parts. Let:
\[ u = x \quad \text{and} \quad dv = e^x \, dx \]
Then:
\[ du = dx \quad \text{and} \quad v = e^x \]

Using the formula for integration by parts:
\[ \int u \, dv = uv - \int v \, du \]

Substitute:
\[ \int x e^x \, dx = x e^x - \int e^x \, dx \]
\[ = x e^x - e^x + C \]
\[ = e^x (x - 1) + C \]

Answer:
\[ \boxed{e^x (x - 1) + C} \]

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(2) \( \int x \sin x \, dx \)

Solution:
We use integration by parts. Let:
\[ u = x \quad \text{and} \quad dv = \sin x \, dx \]
Then:
\[ du = dx \quad \text{and} \quad v = -\cos x \]

Using the formula for integration by parts:
\[ \int u \, dv = uv - \int v \, du \]

Substitute:
\[ \int x \sin x \, dx = x (-\cos x) - \int (-\cos x) \, dx \]
\[ = -x \cos x + \int \cos x \, dx \]
\[ = -x \cos x + \sin x + C \]

Answer:
\[ \boxed{-x \cos x + \sin x + C} \]

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(3) \( \int x \log x \, dx \)

Solution:
We use integration by parts. Let:
\[ u = \log x \quad \text{and} \quad dv = x \, dx \]
Then:
\[ du = \frac{1}{x} \, dx \quad \text{and} \quad v = \frac{x^2}{2} \]

Using the formula for integration by parts:
\[ \int u \, dv = uv - \int v \, du \]

Substitute:
\[ \int x \log x \, dx = \log x \cdot \frac{x^2}{2} - \int \frac{x^2}{2} \cdot \frac{1}{x} \, dx \]
\[ = \frac{x^2 \log x}{2} - \int \frac{x}{2} \, dx \]
\[ = \frac{x^2 \log x}{2} - \frac{1}{2} \int x \, dx \]
\[ = \frac{x^2 \log x}{2} - \frac{1}{2} \cdot \frac{x^2}{2} + C \]
\[ = \frac{x^2 \log x}{2} - \frac{x^2}{4} + C \]

Answer:
\[ \boxed{\frac{x^2 \log x}{2} - \frac{x^2}{4} + C} \]

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(4) \( \int x \sec^2 x \, dx \)

Solution:
We use integration by parts. Let:
\[ u = x \quad \text{and} \quad dv = \sec^2 x \, dx \]
Then:
\[ du = dx \quad \text{and} \quad v = \tan x \]

Using the formula for integration by parts:
\[ \int u \, dv = uv - \int v \, du \]

Substitute:
\[ \int x \sec^2 x \, dx = x \tan x - \int \tan x \, dx \]
\[ = x \tan x - \int \frac{\sin x}{\cos x} \, dx \]
\[ = x \tan x + \log |\cos x| + C \]

Answer:
\[ \boxed{x \tan x + \log |\cos x| + C} \]

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(5) \( \int x \tan^{-1} x \, dx \)

Solution:
We use integration by parts. Let:
\[ u = \tan^{-1} x \quad \text{and} \quad dv = x \, dx \]
Then:
\[ du = \frac{1}{1 + x^2} \, dx \quad \text{and} \quad v = \frac{x^2}{2} \]

Using the formula for integration by parts:
\[ \int u \, dv = uv - \int v \, du \]

Substitute:
\[ \int x \tan^{-1} x \, dx = \tan^{-1} x \cdot \frac{x^2}{2} - \int \frac{x^2}{2} \cdot \frac{1}{1 + x^2} \, dx \]
\[ = \frac{x^2 \tan^{-1} x}{2} - \frac{1}{2} \int \frac{x^2}{1 + x^2} \, dx \]

Simplify the remaining integral:
\[ \frac{x^2}{1 + x^2} = 1 - \frac{1}{1 + x^2} \]
\[ \int \frac{x^2}{1 + x^2} \, dx = \int \left( 1 - \frac{1}{1 + x^2} \right) \, dx \]
\[ = \int 1 \, dx - \int \frac{1}{1 + x^2} \, dx \]
\[ = x - \tan^{-1} x \]

Substitute back:
\[ \int x \tan^{-1} x \, dx = \frac{x^2 \tan^{-1} x}{2} - \frac{1}{2} \left( x - \tan^{-1} x \right) + C \]
\[ = \frac{x^2 \tan^{-1} x}{2} - \frac{x}{2} + \frac{\tan^{-1} x}{2} + C \]
\[ = \frac{x^2 \tan^{-1} x + \tan^{-1} x - x}{2} + C \]
\[ = \frac{(x^2 + 1) \tan^{-1} x - x}{2} + C \]

Answer:
\[ \boxed{\frac{(x^2 + 1) \tan^{-1} x - x}{2} + C} \]

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(6) \( \int \log x \, dx \)

Solution:
We use integration by parts. Let:
\[ u = \log x \quad \text{and} \quad dv = dx \]
Then:
\[ du = \frac{1}{x} \, dx \quad \text{and} \quad v = x \]

Using the formula for integration by parts:
\[ \int u \, dv = uv - \int v \, du \]

Substitute:
\[ \int \log x \, dx = x \log x - \int x \cdot \frac{1}{x} \, dx \]
\[ = x \log x - \int 1 \, dx \]
\[ = x \log x - x + C \]

Answer:
\[ \boxed{x \log x - x + C} \]

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(7) \( \int \sin^{-1} x \, dx \)

Solution:
We use integration by parts. Let:
\[ u = \sin^{-1} x \quad \text{and} \quad dv = dx \]
Then:
\[ du = \frac{1}{\sqrt{1 - x^2}} \, dx \quad \text{and} \quad v = x \]

Using the formula for integration by parts:
\[ \int u \, dv = uv - \int v \, du \]

Substitute:
\[ \int \sin^{-1} x \, dx = x \sin^{-1} x - \int x \cdot \frac{1}{\sqrt{1 - x^2}} \, dx \]

For the remaining integral, let \( w = 1 - x^2 \), so \( dw = -2x \, dx \):
\[ \int \frac{x}{\sqrt{1 - x^2}} \, dx = -\frac{1}{2} \int \frac{1}{\sqrt{w}} \, dw \]
\[ = -\frac{1}{2} \cdot 2\sqrt{w} + C \]
\[ = -\sqrt{1 - x^2} + C \]

Substitute back:
\[ \int \sin^{-1} x \, dx = x \sin^{-1} x + \sqrt{1 - x^2} + C \]

Answer:
\[ \boxed{x \sin^{-1} x + \sqrt{1 - x^2} + C} \]

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(8) \( \int x \sin^2 x \, dx \)

Solution:
Use the identity \( \sin^2 x = \frac{1 - \cos 2x}{2} \):
\[ \int x \sin^2 x \, dx = \int x \cdot \frac{1 - \cos 2x}{2} \, dx \]
\[ = \frac{1}{2} \int x \, dx - \frac{1}{2} \int x \cos 2x \, dx \]

The first integral is straightforward:
\[ \frac{1}{2} \int x \, dx = \frac{x^2}{4} \]

For the second integral, use integration by parts. Let:
\[ u = x \quad \text{and} \quad dv = \cos 2x \, dx \]
Then:
\[ du = dx \quad \text{and} \quad v = \frac{\sin 2x}{2} \]

Using the formula for integration by parts:
\[ \int u \, dv = uv - \int v \, du \]

Substitute:
\[ \int x \cos 2x \, dx = x \cdot \frac{\sin 2x}{2} - \int \frac{\sin 2x}{2} \, dx \]
\[ = \frac{x \sin 2x}{2} - \frac{1}{2} \int \sin 2x \, dx \]
\[ = \frac{x \sin 2x}{2} - \frac{1}{2} \left( -\frac{\cos 2x}{2} \right) \]
\[ = \frac{x \sin 2x}{2} + \frac{\cos 2x}{4} \]

Combine results:
\[ \int x \sin^2 x \, dx = \frac{x^2}{4} - \frac{1}{2} \left( \frac{x \sin 2x}{2} + \frac{\cos 2x}{4} \right) + C \]
\[ = \frac{x^2}{4} - \frac{x \sin 2x}{4} - \frac{\cos 2x}{8} + C \]

Answer:
\[ \boxed{\frac{x^2}{4} - \frac{x \sin 2x}{4} - \frac{\cos 2x}{8} + C} \]

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(9) \( \int x \sin 3x \cos 2x \, dx \)

Solution:
Use the product-to-sum identity:
\[ \sin A \cos B = \frac{1}{2} [\sin(A + B) + \sin(A - B)] \]
Here, \( A = 3x \) and \( B = 2x \):
\[ \sin 3x \cos 2x = \frac{1}{2} [\sin(5x) + \sin(x)] \]

So:
\[ \int x \sin 3x \cos 2x \, dx = \frac{1}{2} \int x [\sin(5x) + \sin(x)] \, dx \]
\[ = \frac{1}{2} \left( \int x \sin(5x) \, dx + \int x \sin(x) \, dx \right) \]

#### First integral: \( \int x \sin(5x) \, dx \)
Use integration by parts. Let:
\[ u = x \quad \text{and} \quad dv = \sin(5x) \, dx \]
Then:
\[ du = dx \quad \text{and} \quad v = -\frac{\cos(5x)}{5} \]

Using the formula for integration by parts:
\[ \int u \, dv = uv - \int v \, du \]

Substitute:
\[ \int x \sin(5x) \, dx = x \left( -\frac{\cos(5x)}{5} \right) - \int -\frac{\cos(5x)}{5} \, dx \]
\[ = -\frac{x \cos(5x)}{5} + \frac{1}{5} \int \cos(5x) \, dx \]
\[ = -\frac{x \cos(5x)}{5} + \frac{1}{5} \cdot \frac{\sin(5x)}{5} \]
\[ = -\frac{x \cos(5x)}{5} + \frac{\sin(5x)}{25} \]

#### Second integral: \( \int x \sin(x) \, dx \)
From earlier (integral 2):
\[ \int x \sin(x) \, dx = -x \cos(x) + \sin(x) \]

Combine results:
\[ \int x \sin 3x \cos 2x \, dx = \frac{1}{2} \left( -\frac{x \cos(5x)}{5} + \frac{\sin(5x)}{25} - x \cos(x) + \sin(x) \right) + C \]
\[ = -\frac{x \cos(5x)}{10} + \frac{\sin(5x)}{50} - \frac{x \cos(x)}{2} + \frac{\sin(x)}{2} + C \]

Answer:
\[ \boxed{-\frac{x \cos(5x)}{10} + \frac{\sin(5x)}{50} - \frac{x \cos(x)}{2} + \frac{\sin(x)}{2} + C} \]

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(10) \( \int x^3 e^{x^2} \, dx \)

Solution:
Let \( u = x^2 \). Then \( du = 2x \, dx \), so \( x \, dx = \frac{du}{2} \). Also, \( x^3 = x^2 \cdot x = u \cdot \sqrt{u} \).

Rewrite the integral:
\[ \int x^3 e^{x^2} \, dx = \int u \cdot e^u \cdot \frac{du}{2} \]
\[ = \frac{1}{2} \int u e^u \, du \]

Now, use integration by parts. Let:
\[ v = u \quad \text{and} \quad dw = e^u \, du \]
Then:
\[ dv = du \quad \text{and} \quad w = e^u \]

Using the formula for integration by parts:
\[ \int v \, dw = vw - \int w \, dv \]

Substitute:
\[ \int u e^u \, du = u e^u - \int e^u \, du \]
\[ = u e^u - e^u + C \]
\[ = e^u (u - 1) + C \]

Substitute back \( u = x^2 \):
\[ \int x^3 e^{x^2} \, dx = \frac{1}{2} e^{x^2} (x^2 - 1) + C \]

Answer:
\[ \boxed{\frac{1}{2} e^{x^2} (x^2 - 1) + C} \]

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Final Answers:

1. \( \boxed{e^x (x - 1) + C} \)
2. \( \boxed{-x \cos x + \sin x + C} \)
3. \( \boxed{\frac{x^2 \log x}{2} - \frac{x^2}{4} + C} \)
4. \( \boxed{x \tan x + \log |\cos x| + C} \)
5. \( \boxed{\frac{(x^2 + 1) \tan^{-1} x - x}{2} + C} \)
6. \( \boxed{x \log x - x + C} \)
7. \( \boxed{x \sin^{-1} x + \sqrt{1 - x^2} + C} \)
8. \( \boxed{\frac{x^2}{4} - \frac{x \sin 2x}{4} - \frac{\cos 2x}{8} + C} \)
9. \( \boxed{-\frac{x \cos(5x)}{10} + \frac{\sin(5x)}{50} - \frac{x \cos(x)}{2} + \frac{\sin(x)}{2} + C} \)
10. \( \boxed{\frac{1}{2} e^{x^2} (x^2 - 1) + C} \)