Calculus Examples

$\int \ln (2 x + 1) d x$

Step 1

Let $u = 2 x + 1$ . Then $d u = 2 d x$ , so $\frac{1}{2} d u = d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 1.1

Let $u = 2 x + 1$ . Find $\frac{d u}{d x}$ .

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Step 1.1.1

Differentiate $2 x + 1$ .

$\frac{d}{d x} [2 x + 1]$

Step 1.1.2

By the Sum Rule, the derivative of $2 x + 1$ with respect to $x$ is $\frac{d}{d x} [2 x] + \frac{d}{d x} [1]$ .

$\frac{d}{d x} [2 x] + \frac{d}{d x} [1]$

Step 1.1.3

Evaluate $\frac{d}{d x} [2 x]$ .

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Step 1.1.3.1

Since $2$ is constant with respect to $x$ , the derivative of $2 x$ with respect to $x$ is $2 \frac{d}{d x} [x]$ .

$2 \frac{d}{d x} [x] + \frac{d}{d x} [1]$

Step 1.1.3.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$2 \cdot 1 + \frac{d}{d x} [1]$

Step 1.1.3.3

Multiply $2$ by $1$ .

$2 + \frac{d}{d x} [1]$

Step 1.1.4

Differentiate using the Constant Rule.

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Step 1.1.4.1

Since $1$ is constant with respect to $x$ , the derivative of $1$ with respect to $x$ is $0$ .

$2 + 0$

Step 1.1.4.2

Add $2$ and $0$ .

$2$

Step 1.2

Rewrite the problem using $u$ and $d u$ .

$\int \ln (u) \frac{1}{2} d u$

Step 2

Combine $\ln (u)$ and $\frac{1}{2}$ .

$\int \frac{\ln (u)}{2} d u$

Step 3

Since $\frac{1}{2}$ is constant with respect to $u$ , move $\frac{1}{2}$ out of the integral.

$\frac{1}{2} \int \ln (u) d u$

Step 4

Integrate by parts using the formula $\int w d v = w v - \int v d w$ , where $w = \ln (u)$ and $dv = 1$ .

$\frac{1}{2} (\ln (u) u - \int u \frac{1}{u} d u)$

Step 5

Simplify.

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Step 5.1

Combine $u$ and $\frac{1}{u}$ .

$\frac{1}{2} (\ln (u) u - \int \frac{u}{u} d u)$

Step 5.2

Cancel the common factor of $u$ .

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Step 5.2.1

Cancel the common factor.

$\frac{1}{2} (\ln (u) u - \int \frac{u}{u} d u)$

Step 5.2.2

Rewrite the expression.

$\frac{1}{2} (\ln (u) u - \int d u)$

Step 6

Apply the constant rule.

$\frac{1}{2} (\ln (u) u - (u + C))$

Step 7

Simplify.

$\frac{1}{2} (\ln (u) u - u) + C$

Step 8

Replace all occurrences of $u$ with $2 x + 1$ .

$\frac{1}{2} (\ln (2 x + 1) (2 x + 1) - (2 x + 1)) + C$

Step 9

Simplify.

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Step 9.1

Simplify each term.

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Step 9.1.1

Apply the distributive property.

$\frac{1}{2} (\ln (2 x + 1) (2 x + 1) - (2 x) - 1 \cdot 1) + C$

Step 9.1.2

Multiply $2$ by $- 1$ .

$\frac{1}{2} (\ln (2 x + 1) (2 x + 1) - 2 x - 1 \cdot 1) + C$

Step 9.1.3

Multiply $- 1$ by $1$ .

$\frac{1}{2} (\ln (2 x + 1) (2 x + 1) - 2 x - 1) + C$

Step 9.2

Apply the distributive property.

$\frac{1}{2} (\ln (2 x + 1) (2 x + 1)) + \frac{1}{2} (- 2 x) + \frac{1}{2} \cdot - 1 + C$

Step 9.3

Simplify.

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Step 9.3.1

Combine $\ln (2 x + 1)$ and $\frac{1}{2}$ .

$\frac{\ln (2 x + 1)}{2} (2 x + 1) + \frac{1}{2} (- 2 x) + \frac{1}{2} \cdot - 1 + C$

Step 9.3.2

Cancel the common factor of $2$ .

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Step 9.3.2.1

Factor $2$ out of $- 2 x$ .

$\frac{\ln (2 x + 1)}{2} (2 x + 1) + \frac{1}{2} (2 (- x)) + \frac{1}{2} \cdot - 1 + C$

Step 9.3.2.2

Cancel the common factor.

$\frac{\ln (2 x + 1)}{2} (2 x + 1) + \frac{1}{2} (2 (- x)) + \frac{1}{2} \cdot - 1 + C$

Step 9.3.2.3

Rewrite the expression.

$\frac{\ln (2 x + 1)}{2} (2 x + 1) - x + \frac{1}{2} \cdot - 1 + C$

Step 9.3.3

Combine $\frac{1}{2}$ and $- 1$ .

$\frac{\ln (2 x + 1)}{2} (2 x + 1) - x + \frac{- 1}{2} + C$

Step 9.4

Simplify each term.

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Step 9.4.1

Apply the distributive property.

$\frac{\ln (2 x + 1)}{2} (2 x) + \frac{\ln (2 x + 1)}{2} \cdot 1 - x + \frac{- 1}{2} + C$

Step 9.4.2

Rewrite using the commutative property of multiplication.

$2 \frac{\ln (2 x + 1)}{2} x + \frac{\ln (2 x + 1)}{2} \cdot 1 - x + \frac{- 1}{2} + C$

Step 9.4.3

Multiply $\frac{\ln (2 x + 1)}{2}$ by $1$ .

$2 \frac{\ln (2 x + 1)}{2} x + \frac{\ln (2 x + 1)}{2} - x + \frac{- 1}{2} + C$

Step 9.4.4

Cancel the common factor of $2$ .

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Step 9.4.4.1

Cancel the common factor.

$2 \frac{\ln (2 x + 1)}{2} x + \frac{\ln (2 x + 1)}{2} - x + \frac{- 1}{2} + C$

Step 9.4.4.2

Rewrite the expression.

$\ln (2 x + 1) x + \frac{\ln (2 x + 1)}{2} - x + \frac{- 1}{2} + C$

Step 9.4.5

To write $\ln (2 x + 1) x$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$\ln (2 x + 1) x \cdot \frac{2}{2} + \frac{\ln (2 x + 1)}{2} - x + \frac{- 1}{2} + C$

Step 9.4.6

Combine $\ln (2 x + 1) x$ and $\frac{2}{2}$ .

$\frac{\ln (2 x + 1) x \cdot 2}{2} + \frac{\ln (2 x + 1)}{2} - x + \frac{- 1}{2} + C$

Step 9.4.7

Combine the numerators over the common denominator.

$\frac{\ln (2 x + 1) x \cdot 2 + \ln (2 x + 1)}{2} - x + \frac{- 1}{2} + C$

Step 9.4.8

Simplify the numerator.

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Step 9.4.8.1

Factor $\ln (2 x + 1)$ out of $\ln (2 x + 1) x \cdot 2 + \ln (2 x + 1)$ .

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Step 9.4.8.1.1

Factor $\ln (2 x + 1)$ out of $\ln (2 x + 1) x \cdot 2$ .

$\frac{\ln (2 x + 1) (x \cdot 2) + \ln (2 x + 1)}{2} - x + \frac{- 1}{2} + C$

Step 9.4.8.1.2

Multiply by $1$ .

$\frac{\ln (2 x + 1) (x \cdot 2) + \ln (2 x + 1) \cdot 1}{2} - x + \frac{- 1}{2} + C$

Step 9.4.8.1.3

Factor $\ln (2 x + 1)$ out of $\ln (2 x + 1) (x \cdot 2) + \ln (2 x + 1) \cdot 1$ .

$\frac{\ln (2 x + 1) (x \cdot 2 + 1)}{2} - x + \frac{- 1}{2} + C$

Step 9.4.8.2

Move $2$ to the left of $x$ .

$\frac{\ln (2 x + 1) (2 x + 1)}{2} - x + \frac{- 1}{2} + C$

Step 9.4.9

Move the negative in front of the fraction.

$\frac{\ln (2 x + 1) (2 x + 1)}{2} - x - \frac{1}{2} + C$

Step 9.5

Combine the numerators over the common denominator.

$- x + \frac{\ln (2 x + 1) (2 x + 1) - 1}{2} + C$

Step 10

Reorder terms.

$- x + \frac{1}{2} (\ln (2 x + 1) (2 x + 1) - 1) + C$