Question Statement

Evaluate the integral:

$\int ln (2 x + 3) d x$

Background and Explanation

This problem demonstrates integration by parts, the standard technique for integrating logarithmic functions. The key insight is to treat $ln (2 x + 3)$ as one part and multiply by $1$ (the identity) to create a product suitable for the integration by parts formula.

Solution

We begin by rewriting the integral to prepare for integration by parts:

$I = \int ln (2 x + 3) \cdot 1 d x$

Step 1: Apply Integration by Parts

Using the formula $\int u d v = uv - \int v d u$ , we choose:

$u = ln (2 x + 3)$ , which gives $d u = \frac{2}{2 x + 3} d x$
$d v = 1 d x$ , which gives $v = x$

Substituting these into the integration by parts formula:

$I = ln (2 x + 3) \cdot \int 1 d x - \int (\int 1 d x) \times \frac{d}{d x} (ln (2 x + 3)) d x = ln (2 x + 3) \cdot x - \int x \times \frac{1}{2 x + 3} \cdot 2 d x = x ln (2 x + 3) - \int \frac{2 x}{2 x + 3} d x$

Step 2: Simplify the Remaining Integral

To evaluate $\int \frac{2 x}{2 x + 3} d x$ , we manipulate the numerator by adding and subtracting 3 to create a simpler expression:

$I = x ln (2 x + 3) - \int \frac{2 x + 3 - 3}{2 x + 3} d x = x ln (2 x + 3) - \int (\frac{2 x + 3}{2 x + 3} - \frac{3}{2 x + 3}) d x = x ln (2 x + 3) - \int 1 d x + 3 \int \frac{1}{2 x + 3} d x$

Step 3: Evaluate the Final Integrals

The first integral is straightforward: $\int 1 d x = x$ .

For the second integral, we adjust the coefficient to match the derivative of the denominator $(2 x + 3)^{'} = 2$ :

$= x ln (2 x + 3) - x + \frac{3}{2} \int \frac{2}{2 x + 3} d x$

Using the logarithmic integration formula $\int \frac{f ^{'} ( x )}{f ( x )} d x = ln f (x)$ :

$= x ln (2 x + 3) - x + \frac{3}{2} ln (2 x + 3) + c$

Key Formulas or Methods Used

Integration by Parts: $\int u d v = uv - \int v d u$
Algebraic Manipulation: Adding and subtracting constants in the numerator to decompose rational functions ( $\frac{2 x}{2 x + 3} = \frac{( 2 x + 3 ) - 3}{2 x + 3}$ )
Logarithmic Integration: $\int \frac{f ^{'} ( x )}{f ( x )} d x = ln ∣ f (x) ∣ + c$ (applied here as $\int \frac{2}{2 x + 3} d x = ln (2 x + 3)$ )

Summary of Steps

Rewrite the integral as $\int ln (2 x + 3) \cdot 1 d x$ to prepare for integration by parts
Set $u = ln (2 x + 3)$ and $d v = d x$ , then compute $d u = \frac{2}{2 x + 3} d x$ and $v = x$
Apply integration by parts to obtain $x ln (2 x + 3) - \int \frac{2 x}{2 x + 3} d x$
Rewrite the fraction $\frac{2 x}{2 x + 3}$ as $1 - \frac{3}{2 x + 3}$ by adding and subtracting 3 in the numerator
Split the integral into $\int 1 d x - 3 \int \frac{1}{2 x + 3} d x$ (note the sign change)
Multiply the second integral by $\frac{2}{2}$ to get $\frac{3}{2} \int \frac{2}{2 x + 3} d x$
Evaluate to $\frac{3}{2} ln (2 x + 3)$ and combine all terms with the constant of integration $c$

Question Statement

Evaluate the integral:

$\int ln (2 x + 3) d x$

Background and Explanation

Solution

We begin by rewriting the integral to prepare for integration by parts:

$I = \int ln (2 x + 3) \cdot 1 d x$

Step 1: Apply Integration by Parts

Using the formula $\int u d v = uv - \int v d u$ , we choose:

$u = ln (2 x + 3)$ , which gives $d u = \frac{2}{2 x + 3} d x$
$d v = 1 d x$ , which gives $v = x$

Substituting these into the integration by parts formula:

Step 2: Simplify the Remaining Integral

To evaluate $\int \frac{2 x}{2 x + 3} d x$ , we manipulate the numerator by adding and subtracting 3 to create a simpler expression:

$I = x ln (2 x + 3) - \int \frac{2 x + 3 - 3}{2 x + 3} d x = x ln (2 x + 3) - \int (\frac{2 x + 3}{2 x + 3} - \frac{3}{2 x + 3}) d x = x ln (2 x + 3) - \int 1 d x + 3 \int \frac{1}{2 x + 3} d x$

Step 3: Evaluate the Final Integrals

The first integral is straightforward: $\int 1 d x = x$ .

For the second integral, we adjust the coefficient to match the derivative of the denominator $(2 x + 3)^{'} = 2$ :

$= x ln (2 x + 3) - x + \frac{3}{2} \int \frac{2}{2 x + 3} d x$

Using the logarithmic integration formula $\int \frac{f ^{'} ( x )}{f ( x )} d x = ln f (x)$ :

$= x ln (2 x + 3) - x + \frac{3}{2} ln (2 x + 3) + c$

Key Formulas or Methods Used

Integration by Parts: $\int u d v = uv - \int v d u$
Algebraic Manipulation: Adding and subtracting constants in the numerator to decompose rational functions ( $\frac{2 x}{2 x + 3} = \frac{( 2 x + 3 ) - 3}{2 x + 3}$ )
Logarithmic Integration: $\int \frac{f ^{'} ( x )}{f ( x )} d x = ln ∣ f (x) ∣ + c$ (applied here as $\int \frac{2}{2 x + 3} d x = ln (2 x + 3)$ )

Summary of Steps

Rewrite the integral as $\int ln (2 x + 3) \cdot 1 d x$ to prepare for integration by parts
Set $u = ln (2 x + 3)$ and $d v = d x$ , then compute $d u = \frac{2}{2 x + 3} d x$ and $v = x$
Apply integration by parts to obtain $x ln (2 x + 3) - \int \frac{2 x}{2 x + 3} d x$
Rewrite the fraction $\frac{2 x}{2 x + 3}$ as $1 - \frac{3}{2 x + 3}$ by adding and subtracting 3 in the numerator
Split the integral into $\int 1 d x - 3 \int \frac{1}{2 x + 3} d x$ (note the sign change)
Multiply the second integral by $\frac{2}{2}$ to get $\frac{3}{2} \int \frac{2}{2 x + 3} d x$
Evaluate to $\frac{3}{2} ln (2 x + 3)$ and combine all terms with the constant of integration $c$

3.4 Q-8

Question Statement

Background and Explanation

Solution

Key Formulas or Methods Used

Summary of Steps

3.4 Q-8

Question Statement

Background and Explanation

Solution

Key Formulas or Methods Used

Summary of Steps