Question Statement

Evaluate the indefinite integral:

$\int \frac{2 a x + b}{a x ^{2} + b x + c} d x$

Background and Explanation

This problem demonstrates the substitution method for integration when the numerator is the derivative of the denominator. Recognizing this pattern allows us to simplify the integral to the standard logarithmic form $\int \frac{1}{t} d t$ .

Solution

Let

$I = \int \frac{2 a x + b}{a x ^{2} + b x + c} d x \dots (1)$

We use the method of substitution. Observe that the numerator $(2 a x + b)$ is exactly the derivative of the denominator $(a x^{2} + b x + c)$ .

Put

$t = a x^{2} + b x + c$

Differentiating both sides with respect to $x$ :

$(2 a x + b + 0) d x = d t$

Therefore:

$(2 a x + b) d x = d t$

Substituting into equation (1), we get:

$I = \int \frac{d t}{t}$

This can be written as:

$I = \int \frac{1}{t} d t$

Integrating with respect to $t$ :

$I = ln t + c_{1}$

Back-substituting $t = a x^{2} + b x + c$ :

$I = ln (a x^{2} + b x + c) + c_{1}$

Thus, the solution is:

$\int \frac{2 a x + b}{a x ^{2} + b x + c} d x = ln (a x^{2} + b x + c) + c_{1}$

Key Formulas or Methods Used

Integration by substitution: When $\frac{d}{d x} [f (x)]$ appears as a factor in the integrand, substitute $t = f (x)$
Differential form: $d t = f^{'} (x) d x$ or $\frac{d t}{d x} = f^{'} (x)$
Standard integral: $\int \frac{1}{t} d t = ln ∣ t ∣ + C$ (here applied as $ln t + c_{1}$ assuming $t > 0$ )
Derivative of quadratic: $\frac{d}{d x} (a x^{2} + b x + c) = 2 a x + b$

Summary of Steps

Identify the pattern: Recognize that the numerator $(2 a x + b)$ is the derivative of the denominator $(a x^{2} + b x + c)$
Substitute: Let $t = a x^{2} + b x + c$
Differentiate: Compute $d t = (2 a x + b) d x$
Transform the integral: Replace $(2 a x + b) d x$ with $d t$ and the denominator with $t$ to get $\int \frac{d t}{t}$
Integrate: Apply the logarithmic rule to get $ln t + c_{1}$
Back-substitute: Replace $t$ with $a x^{2} + b x + c$ to obtain the final answer

Question Statement

Evaluate the indefinite integral:

$\int \frac{2 a x + b}{a x ^{2} + b x + c} d x$

Background and Explanation

Solution

Let

$I = \int \frac{2 a x + b}{a x ^{2} + b x + c} d x \dots (1)$

We use the method of substitution. Observe that the numerator $(2 a x + b)$ is exactly the derivative of the denominator $(a x^{2} + b x + c)$ .

Put

$t = a x^{2} + b x + c$

Differentiating both sides with respect to $x$ :

$(2 a x + b + 0) d x = d t$

Therefore:

$(2 a x + b) d x = d t$

Substituting into equation (1), we get:

$I = \int \frac{d t}{t}$

This can be written as:

$I = \int \frac{1}{t} d t$

Integrating with respect to $t$ :

$I = ln t + c_{1}$

Back-substituting $t = a x^{2} + b x + c$ :

$I = ln (a x^{2} + b x + c) + c_{1}$

Thus, the solution is:

$\int \frac{2 a x + b}{a x ^{2} + b x + c} d x = ln (a x^{2} + b x + c) + c_{1}$

Key Formulas or Methods Used

Integration by substitution: When $\frac{d}{d x} [f (x)]$ appears as a factor in the integrand, substitute $t = f (x)$
Differential form: $d t = f^{'} (x) d x$ or $\frac{d t}{d x} = f^{'} (x)$
Standard integral: $\int \frac{1}{t} d t = ln ∣ t ∣ + C$ (here applied as $ln t + c_{1}$ assuming $t > 0$ )
Derivative of quadratic: $\frac{d}{d x} (a x^{2} + b x + c) = 2 a x + b$

Summary of Steps

Identify the pattern: Recognize that the numerator $(2 a x + b)$ is the derivative of the denominator $(a x^{2} + b x + c)$
Substitute: Let $t = a x^{2} + b x + c$
Differentiate: Compute $d t = (2 a x + b) d x$
Transform the integral: Replace $(2 a x + b) d x$ with $d t$ and the denominator with $t$ to get $\int \frac{d t}{t}$
Integrate: Apply the logarithmic rule to get $ln t + c_{1}$
Back-substitute: Replace $t$ with $a x^{2} + b x + c$ to obtain the final answer

3.3 Q-9

Question Statement

Background and Explanation

Solution

Key Formulas or Methods Used

Summary of Steps

3.3 Q-9

Question Statement

Background and Explanation

Solution

Key Formulas or Methods Used

Summary of Steps