U-Substitution for Integration | Formula, Steps & Examples ... - Free Printable

Let's solve each integral step by step.

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1. \( \int (3x + 7)^4 \, dx \)

#### Solution:
We use the substitution method. Let:
\[ u = 3x + 7 \]
Then:
\[ du = 3 \, dx \quad \Rightarrow \quad dx = \frac{1}{3} \, du \]

Substitute \( u \) and \( dx \) into the integral:
\[ \int (3x + 7)^4 \, dx = \int u^4 \cdot \frac{1}{3} \, du = \frac{1}{3} \int u^4 \, du \]

Now, integrate \( u^4 \):
\[ \frac{1}{3} \int u^4 \, du = \frac{1}{3} \cdot \frac{u^5}{5} + C = \frac{u^5}{15} + C \]

Substitute back \( u = 3x + 7 \):
\[ \frac{u^5}{15} + C = \frac{(3x + 7)^5}{15} + C \]

#### Final Answer:
\[ \boxed{\frac{(3x + 7)^5}{15} + C} \]

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2. \( \int 3x \sqrt{x^2 - 5} \, dx \)

#### Solution:
We use the substitution method. Let:
\[ u = x^2 - 5 \]
Then:
\[ du = 2x \, dx \quad \Rightarrow \quad x \, dx = \frac{1}{2} \, du \]

Substitute \( u \) and \( x \, dx \) into the integral:
\[ \int 3x \sqrt{x^2 - 5} \, dx = \int 3 \cdot \sqrt{u} \cdot \frac{1}{2} \, du = \frac{3}{2} \int \sqrt{u} \, du \]

Now, integrate \( \sqrt{u} \):
\[ \frac{3}{2} \int \sqrt{u} \, du = \frac{3}{2} \int u^{1/2} \, du = \frac{3}{2} \cdot \frac{u^{3/2}}{3/2} + C = u^{3/2} + C \]

Substitute back \( u = x^2 - 5 \):
\[ u^{3/2} + C = (x^2 - 5)^{3/2} + C \]

#### Final Answer:
\[ \boxed{(x^2 - 5)^{3/2} + C} \]

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3. \( \int \sec^2(x) \tan(x) \, dx \)

#### Solution:
We use the substitution method. Let:
\[ u = \tan(x) \]
Then:
\[ du = \sec^2(x) \, dx \]

Substitute \( u \) and \( du \) into the integral:
\[ \int \sec^2(x) \tan(x) \, dx = \int u \, du \]

Now, integrate \( u \):
\[ \int u \, du = \frac{u^2}{2} + C \]

Substitute back \( u = \tan(x) \):
\[ \frac{u^2}{2} + C = \frac{\tan^2(x)}{2} + C \]

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

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4. \( \int e e^{e^x} e^x \, dx \)

#### Solution:
First, simplify the integrand:
\[ e e^{e^x} e^x = e^{1 + e^x + x} \]

So the integral becomes:
\[ \int e^{1 + e^x + x} \, dx \]

We use the substitution method. Let:
\[ u = e^x \]
Then:
\[ du = e^x \, dx \]

Notice that \( e^{1 + e^x + x} = e \cdot e^{e^x} \cdot e^x \). Substituting \( u = e^x \) and \( du = e^x \, dx \):
\[ \int e e^{e^x} e^x \, dx = \int e e^u \, du = e \int e^u \, du \]

Now, integrate \( e^u \):
\[ e \int e^u \, du = e \cdot e^u + C = e e^u + C \]

Substitute back \( u = e^x \):
\[ e e^u + C = e e^{e^x} + C \]

#### Final Answer:
\[ \boxed{e e^{e^x} + C} \]

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5. \( \int \frac{2x \ln(x^2 + 5)}{x^2 + 5} \, dx \)

#### Solution:
We use the substitution method. Let:
\[ u = \ln(x^2 + 5) \]
Then:
\[ du = \frac{1}{x^2 + 5} \cdot 2x \, dx \quad \Rightarrow \quad \frac{2x}{x^2 + 5} \, dx = du \]

Substitute \( u \) and \( du \) into the integral:
\[ \int \frac{2x \ln(x^2 + 5)}{x^2 + 5} \, dx = \int u \, du \]

Now, integrate \( u \):
\[ \int u \, du = \frac{u^2}{2} + C \]

Substitute back \( u = \ln(x^2 + 5) \):
\[ \frac{u^2}{2} + C = \frac{[\ln(x^2 + 5)]^2}{2} + C \]

#### Final Answer:
\[ \boxed{\frac{[\ln(x^2 + 5)]^2}{2} + C} \]

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Summary of Final Answers:

1. \( \boxed{\frac{(3x + 7)^5}{15} + C} \)
2. \( \boxed{(x^2 - 5)^{3/2} + C} \)
3. \( \boxed{\frac{\tan^2(x)}{2} + C} \)
4. \( \boxed{e e^{e^x} + C} \)
5. \( \boxed{\frac{[\ln(x^2 + 5)]^2}{2} + C} \)