Question Statement

Q7. Evaluate the integral: $\int \frac{7 x ^{2} + 7 x + 4}{( 2 x + 1 ) ( x ^{2} + x + 1 )} d x$

Background and Explanation

These problems involve integrating rational functions using partial fraction decomposition. When the degree of the numerator is less than the denominator (proper fraction), we decompose directly. When the degree of the numerator is greater (improper fraction), we first perform polynomial long division to separate the polynomial part from the proper rational remainder.

Solution

Problem 1: $\int \frac{7 x ^{2} + 7 x + 4}{( 2 x + 1 ) ( x ^{2} + x + 1 )} d x$

Let $I = \int \frac{7 x ^{2} + 7 x + 4}{( 2 x + 1 ) ( x ^{2} + x + 1 )} d x$

Consider the partial fraction decomposition: $\frac{7 x ^{2} + 7 x + 4}{( 2 x + 1 ) ( x ^{2} + x + 1 )} = \frac{A}{2 x + 1} + \frac{B x + C}{x ^{2} + x + 1}$

Multiplying both sides by $(2 x + 1) (x^{2} + x + 1)$ : $7 x^{2} + 7 x + 4 = A (x^{2} + x + 1) + (B x + C) (2 x + 1)$

Step 1: Find $A$ using substitution

Put $2 x + 1 = 0 \Rightarrow x = \frac{- 1}{2}$ in equation (2):

$7 (\frac{- 1}{2})^{2} + 7 (\frac{- 1}{2}) + 4 = A ((\frac{- 1}{2})^{2} + (\frac{- 1}{2}) + 1) + (B (\frac{- 1}{2}) + C) (0)$

$7 (\frac{1}{4}) - \frac{7}{2} + 4 = A (\frac{1}{4} - \frac{1}{2} + 1) + 0$

$\frac{7 - 14 + 16}{4} = A (\frac{1 - 2 + 4}{4})$

$\frac{9}{4} = A (\frac{3}{4})$

$A = 3$

Step 2: Find $B$ by comparing coefficients

Comparing coefficients of $x^{2}$ in equation (2):

$7 = A + 2 B$ $7 = 3 + 2 B$ $4 = 2 B$ $B = 2$

Step 3: Find $C$ by comparing coefficients

From the expansion of equation (2): $7 x^{2} + 7 x + 4 = (A + 2 B) x^{2} + (A + B + 2 C) x + (A + C)$

Comparing constant terms: $4 = A + C$ $4 = 3 + C$ $C = 1$

(Note: The integration of the resulting partial fractions $\frac{3}{2 x + 1} + \frac{2 x + 1}{x ^{2} + x + 1}$ would proceed using substitution, but the provided solution transitions to a different problem at this stage.)

Key Formulas or Methods Used

Partial Fraction Decomposition: $\frac{P ( x )}{( a x + b ) ( c x ^{2} + d x + e )} = \frac{A}{a x + b} + \frac{B x + C}{c x ^{2} + d x + e}$ for irreducible quadratic factors
Polynomial Long Division: $\frac{Numerator}{Denominator} = Quotient + \frac{Remainder}{Denominator}$ when degree of numerator $\geq$ degree of denominator
Heaviside Cover-Up Method: Substituting roots of linear factors to find partial fraction constants
Coefficient Comparison: Equating coefficients of like powers of $x$ to solve for unknown constants
Standard Integrals: $\int \frac{1}{x + a} d x = ln ∣ x + a ∣ + C$ and $\int x^{n} d x = \frac{x ^{n + 1}}{n + 1} + C$

Summary of Steps

Set up partial fraction decomposition: $\frac{A}{2 x + 1} + \frac{B x + C}{x ^{2} + x + 1}$
Multiply through by the denominator to clear fractions
Substitute $x = - \frac{1}{2}$ (root of $2 x + 1$ ) to find $A = 3$
Compare $x^{2}$ coefficients to find $B = 2$
Compare constant terms to find $C = 1$
Decomposition yields: $\frac{3}{2 x + 1} + \frac{2 x + 1}{x ^{2} + x + 1}$ (ready for integration)

Question Statement

Q7. Evaluate the integral: $\int \frac{7 x ^{2} + 7 x + 4}{( 2 x + 1 ) ( x ^{2} + x + 1 )} d x$

Background and Explanation

Solution

Problem 1: $\int \frac{7 x ^{2} + 7 x + 4}{( 2 x + 1 ) ( x ^{2} + x + 1 )} d x$

Let $I = \int \frac{7 x ^{2} + 7 x + 4}{( 2 x + 1 ) ( x ^{2} + x + 1 )} d x$

Consider the partial fraction decomposition: $\frac{7 x ^{2} + 7 x + 4}{( 2 x + 1 ) ( x ^{2} + x + 1 )} = \frac{A}{2 x + 1} + \frac{B x + C}{x ^{2} + x + 1}$

Multiplying both sides by $(2 x + 1) (x^{2} + x + 1)$ : $7 x^{2} + 7 x + 4 = A (x^{2} + x + 1) + (B x + C) (2 x + 1)$

Step 1: Find $A$ using substitution

Put $2 x + 1 = 0 \Rightarrow x = \frac{- 1}{2}$ in equation (2):

$7 (\frac{- 1}{2})^{2} + 7 (\frac{- 1}{2}) + 4 = A ((\frac{- 1}{2})^{2} + (\frac{- 1}{2}) + 1) + (B (\frac{- 1}{2}) + C) (0)$

$7 (\frac{1}{4}) - \frac{7}{2} + 4 = A (\frac{1}{4} - \frac{1}{2} + 1) + 0$

$\frac{7 - 14 + 16}{4} = A (\frac{1 - 2 + 4}{4})$

$\frac{9}{4} = A (\frac{3}{4})$

$A = 3$

Step 2: Find $B$ by comparing coefficients

Comparing coefficients of $x^{2}$ in equation (2):

$7 = A + 2 B$ $7 = 3 + 2 B$ $4 = 2 B$ $B = 2$

Step 3: Find $C$ by comparing coefficients

From the expansion of equation (2): $7 x^{2} + 7 x + 4 = (A + 2 B) x^{2} + (A + B + 2 C) x + (A + C)$

Comparing constant terms: $4 = A + C$ $4 = 3 + C$ $C = 1$

Key Formulas or Methods Used

Partial Fraction Decomposition: $\frac{P ( x )}{( a x + b ) ( c x ^{2} + d x + e )} = \frac{A}{a x + b} + \frac{B x + C}{c x ^{2} + d x + e}$ for irreducible quadratic factors
Polynomial Long Division: $\frac{Numerator}{Denominator} = Quotient + \frac{Remainder}{Denominator}$ when degree of numerator $\geq$ degree of denominator
Heaviside Cover-Up Method: Substituting roots of linear factors to find partial fraction constants
Coefficient Comparison: Equating coefficients of like powers of $x$ to solve for unknown constants
Standard Integrals: $\int \frac{1}{x + a} d x = ln ∣ x + a ∣ + C$ and $\int x^{n} d x = \frac{x ^{n + 1}}{n + 1} + C$

Summary of Steps

Set up partial fraction decomposition: $\frac{A}{2 x + 1} + \frac{B x + C}{x ^{2} + x + 1}$
Multiply through by the denominator to clear fractions
Substitute $x = - \frac{1}{2}$ (root of $2 x + 1$ ) to find $A = 3$
Compare $x^{2}$ coefficients to find $B = 2$
Compare constant terms to find $C = 1$
Decomposition yields: $\frac{3}{2 x + 1} + \frac{2 x + 1}{x ^{2} + x + 1}$ (ready for integration)

3.5 Q-7

Question Statement

Background and Explanation

Solution

Problem 1: $\int \frac{7 x ^{2} + 7 x + 4}{( 2 x + 1 ) ( x ^{2} + x + 1 )} d x$

Key Formulas or Methods Used

Summary of Steps

3.5 Q-7

Question Statement

Background and Explanation

Solution

Problem 1: $\int \frac{7 x ^{2} + 7 x + 4}{( 2 x + 1 ) ( x ^{2} + x + 1 )} d x$

Key Formulas or Methods Used

Summary of Steps

Question Statement

Background and Explanation

Solution

Problem 1: ∫(2x+1)(x2+x+1)7x2+7x+4​dx

Key Formulas or Methods Used

Summary of Steps

Question Statement

Background and Explanation

Solution

Problem 1: ∫(2x+1)(x2+x+1)7x2+7x+4​dx

Key Formulas or Methods Used

Summary of Steps

Problem 1: $\int \frac{7 x ^{2} + 7 x + 4}{( 2 x + 1 ) ( x ^{2} + x + 1 )} d x$

Problem 1: $\int \frac{7 x ^{2} + 7 x + 4}{( 2 x + 1 ) ( x ^{2} + x + 1 )} d x$