Samacheer Kalvi Class 12 Business Maths Solutions Chapter 2 Integral Calculus I Exercise 2.2

Question 1. Integrate the following with respect to x. \( [\sqrt { 2x} - \frac { 1 }{\sqrt { 2x}}]² \)
Answer:Let the given integral be \( I \). We have the expression \( (\sqrt{2x} - \frac{1}{\sqrt{2x}})^2 \). Using the identity \( (a-b)^2 = a^2 - 2ab + b^2 \), we get: \( (\sqrt{2x})^2 - 2(\sqrt{2x})(\frac{1}{\sqrt{2x}}) + (\frac{1}{\sqrt{2x}})^2 \) \( = 2x - 2 + \frac{1}{2x} \) Now, we need to integrate this expression with respect to \( x \): \( I = \int (2x - 2 + \frac{1}{2x}) \,dx \) We can integrate each term separately: \( \int 2x \,dx = 2 \int x \,dx = 2 \frac{x^2}{2} = x^2 \) \( \int -2 \,dx = -2x \) \( \int \frac{1}{2x} \,dx = \frac{1}{2} \int \frac{1}{x} \,dx = \frac{1}{2} \log|x| \) Combining these, we get: \( I = x^2 - 2x + \frac{1}{2} \log|x| + c \) Here, \( c \) is the constant of integration, which is always added after indefinite integration.
In simple words: First, expand the given expression using the square formula. Then, integrate each part separately. The integral of \( x \) is \( x^2/2 \), the integral of a constant is constant times \( x \), and the integral of \( 1/x \) is \( \log|x| \). Remember to add 'c' at the end.

🎯 Exam Tip: Always expand algebraic expressions before integrating, especially if they are powers or products, to simplify the integration process.

Question 2. \( \frac {x^4+x^2+2 }{(x-1)} \)
Answer:Let the given integral be \( I \). We need to integrate \( \frac{x^4+x^2+2}{x-1} \) with respect to \( x \). Since the degree of the numerator (4) is greater than the degree of the denominator (1), we perform polynomial long division or algebraic manipulation. We can rewrite the numerator: \( x^4+x^2+2 = x^4 - 1 + x^2 + 1 + 2 = (x^2-1)(x^2+1) + x^2 + 3 \) This is one way to approach it. A more direct method is polynomial division or trying to make the denominator appear in the numerator. Let's use algebraic manipulation by adding and subtracting terms to match \( x-1 \): \( \frac{x^4+x^2+2}{x-1} = \frac{x^4 - x^3 + x^3 - x^2 + 2x^2 - 2x + 2x + 2}{x-1} \) \( = \frac{x^3(x-1) + x^2(x-1) + 2x(x-1) + 2x+2}{x-1} \) \( = x^3 + x^2 + 2x + \frac{2x+2}{x-1} \) Now we need to integrate \( \int (x^3 + x^2 + 2x + \frac{2x+2}{x-1}) \,dx \) Let's simplify \( \frac{2x+2}{x-1} \): \( \frac{2x+2}{x-1} = \frac{2(x-1)+4}{x-1} = 2 + \frac{4}{x-1} \) So the integral becomes: \( I = \int (x^3 + x^2 + 2x + 2 + \frac{4}{x-1}) \,dx \) Now, integrate each term: \( \int x^3 \,dx = \frac{x^4}{4} \) \( \int x^2 \,dx = \frac{x^3}{3} \) \( \int 2x \,dx = \frac{2x^2}{2} = x^2 \) \( \int 2 \,dx = 2x \) \( \int \frac{4}{x-1} \,dx = 4 \log|x-1| \) Combining these results: \( I = \frac{x^4}{4} + \frac{x^3}{3} + x^2 + 2x + 4 \log|x-1| + c \) The source solution uses slightly different initial manipulation: \( \int \frac{x^4-x^2+2}{x-1} \,dx = \int (\frac{x^2(x^2-1)}{x-1} + \frac{2}{x-1}) \,dx \) (The original question had \( x^4+x^2+2 \), the OCR appears to have transcribed it as \( x^4-x^2+2 \) in the solution's first step. We will follow the question as written.) Let's perform polynomial long division for \( \frac{x^4+x^2+2}{x-1} \): \( (x^4+x^2+2) \div (x-1) \) \( x^3(x-1) = x^4-x^3 \), remainder: \( x^3+x^2+2 \) \( x^2(x-1) = x^3-x^2 \), remainder: \( 2x^2+2 \) \( 2x(x-1) = 2x^2-2x \), remainder: \( 2x+2 \) \( 2(x-1) = 2x-2 \), remainder: \( 4 \) So, \( \frac{x^4+x^2+2}{x-1} = x^3+x^2+2x+2 + \frac{4}{x-1} \) Integrating this: \( I = \int (x^3 + x^2 + 2x + 2 + \frac{4}{x-1}) \,dx \) \( I = \frac{x^4}{4} + \frac{x^3}{3} + \frac{2x^2}{2} + 2x + 4 \log|x-1| + c \) \( I = \frac{x^4}{4} + \frac{x^3}{3} + x^2 + 2x + 4 \log|x-1| + c \) The solution provided in the image starts with \( \int \frac{x^4-x^2+2}{x-1} dx \). Let's use the question provided: \( x^4+x^2+2 \). The source solution's steps: \( \int \frac{x^4-x^2+2}{x-1} \,dx = \int (\frac{x^2(x^2-1)}{x-1} + \frac{2}{x-1}) \,dx \) \( = \int (\frac{x^2(x-1)(x+1)}{x-1} + \frac{2}{x-1}) \,dx \) \( = \int (x^2(x+1) + \frac{2}{x-1}) \,dx \) \( = \int (x^3+x^2 + \frac{2}{x-1}) \,dx \) \( = \frac{x^4}{4} + \frac{x^3}{3} + 2 \log|x-1| + c \) This solution follows a slightly different numerator (using \( x^4-x^2+2 \) instead of \( x^4+x^2+2 \)). Given the strict instruction to follow the question text exactly, and the answer's first step seems to differ, I will provide the solution based on the question as written. Let's simplify the original fraction using long division: \( \frac{x^4+x^2+2}{x-1} \) Divide \( x^4+x^2+2 \) by \( x-1 \): Quotient: \( x^3+x^2+2x+2 \) Remainder: \( 4 \) So, \( \frac{x^4+x^2+2}{x-1} = x^3+x^2+2x+2 + \frac{4}{x-1} \) Now, integrate this expression: \( \int (x^3+x^2+2x+2 + \frac{4}{x-1}) \,dx \) \( = \frac{x^4}{4} + \frac{x^3}{3} + \frac{2x^2}{2} + 2x + 4 \log|x-1| + c \) \( = \frac{x^4}{4} + \frac{x^3}{3} + x^2 + 2x + 4 \log|x-1| + c \) This matches the correct integration of the given problem statement. The original source's solution uses a modified numerator \( (x^4-x^2+2) \) in its first step, leading to a different result. We stick to the question as stated.
In simple words: When the top part of the fraction has a higher power of \( x \) than the bottom part, first divide the top by the bottom. This changes the fraction into easier parts. Then, integrate each part separately. Remember that \( \int \frac{1}{ax+b} dx = \frac{1}{a} \log|ax+b| \). Finally, add the constant of integration.

🎯 Exam Tip: Always perform polynomial long division if the degree of the numerator is greater than or equal to the degree of the denominator before attempting to integrate rational functions.

Question 3. \( \frac {x^3 }{x+2} \)
Answer:Let the given integral be \( I \). We need to integrate \( \frac{x^3}{x+2} \) with respect to \( x \). Again, the degree of the numerator (3) is greater than the degree of the denominator (1), so we perform polynomial long division or algebraic manipulation. We can add and subtract 8 in the numerator to use the sum/difference of cubes formula: \( x^3 = x^3 + 8 - 8 = (x+2)(x^2-2x+4) - 8 \) So, \( \frac{x^3}{x+2} = \frac{(x+2)(x^2-2x+4) - 8}{x+2} \) \( = x^2-2x+4 - \frac{8}{x+2} \) Now, we need to integrate this expression: \( I = \int (x^2-2x+4 - \frac{8}{x+2}) \,dx \) Integrate each term: \( \int x^2 \,dx = \frac{x^3}{3} \) \( \int -2x \,dx = -\frac{2x^2}{2} = -x^2 \) \( \int 4 \,dx = 4x \) \( \int -\frac{8}{x+2} \,dx = -8 \log|x+2| \) Combining these, we get: \( I = \frac{x^3}{3} - x^2 + 4x - 8 \log|x+2| + c \) The constant of integration \( c \) is added at the end. This method uses a clever algebraic trick to simplify the expression before integration.
In simple words: To integrate this fraction, we first rewrite the top part so that the bottom part \((x+2)\) can be factored out. We use the formula for a cube \(x^3+a^3\). Then, we split the fraction into simpler terms. Finally, we integrate each simple term. The integral of \(1/x\) is \( \log|x| \).

🎯 Exam Tip: For rational functions where the numerator's degree is higher, consider using polynomial long division or specific algebraic identities (like \(a^3+b^3\) or \(a^3-b^3\)) to simplify the integrand before integration.

Question 4. \( \frac {x^3+3x^2-7x+11 }{x+5} \)
Answer:Let the given integral be \( I \). We need to integrate \( \frac{x^3+3x^2-7x+11}{x+5} \) with respect to \( x \). Since the degree of the numerator (3) is greater than the degree of the denominator (1), we perform polynomial long division. Let's divide \( x^3+3x^2-7x+11 \) by \( x+5 \): x^2 - 2x + 3 ________________ x + 5 | x^3 + 3x^2 - 7x + 11 -(x^3 + 5x^2) ___________ -2x^2 - 7x -(-2x^2 - 10x) ___________ 3x + 11 -(3x + 15) ________ -4 So, \( \frac{x^3+3x^2-7x+11}{x+5} = x^2-2x+3 - \frac{4}{x+5} \) Now, we integrate this expression: \( I = \int (x^2-2x+3 - \frac{4}{x+5}) \,dx \) Integrate each term separately: \( \int x^2 \,dx = \frac{x^3}{3} \) \( \int -2x \,dx = -\frac{2x^2}{2} = -x^2 \) \( \int 3 \,dx = 3x \) \( \int -\frac{4}{x+5} \,dx = -4 \log|x+5| \) Combining these, we get the final result: \( I = \frac{x^3}{3} - x^2 + 3x - 4 \log|x+5| + c \) The result includes the constant of integration, \( c \).
In simple words: When the top polynomial has a higher power than the bottom, divide the top by the bottom using long division. This breaks the complicated fraction into a simpler polynomial and a smaller fraction. Then, integrate each part of the result. The integral of \( 1/(x+a) \) is \( \log|x+a| \).

🎯 Exam Tip: Long division is a reliable method for integrating rational functions where the numerator's degree is greater than or equal to the denominator's degree. Perform it carefully to avoid calculation errors.

Question 5. \( \frac { 3x+2 }{(x-2)(x-3)} \)
Answer:Let the given integral be \( I \). We need to integrate \( \frac{3x+2}{(x-2)(x-3)} \) with respect to \( x \). Since the denominator is a product of distinct linear factors, we use partial fraction decomposition. We write \( \frac{3x+2}{(x-2)(x-3)} = \frac{A}{x-2} + \frac{B}{x-3} \) To find \( A \) and \( B \), we multiply both sides by \( (x-2)(x-3) \): \( 3x+2 = A(x-3) + B(x-2) \) To find \( A \), set \( x=2 \): \( 3(2)+2 = A(2-3) + B(2-2) \) \( 6+2 = A(-1) + B(0) \) \( 8 = -A \implies A = -8 \) To find \( B \), set \( x=3 \): \( 3(3)+2 = A(3-3) + B(3-2) \) \( 9+2 = A(0) + B(1) \) \( 11 = B \implies B = 11 \) So, the partial fraction decomposition is: \( \frac{3x+2}{(x-2)(x-3)} = \frac{-8}{x-2} + \frac{11}{x-3} \) Now, integrate this expression: \( I = \int (\frac{-8}{x-2} + \frac{11}{x-3}) \,dx \) Integrate each term: \( \int \frac{-8}{x-2} \,dx = -8 \log|x-2| \) \( \int \frac{11}{x-3} \,dx = 11 \log|x-3| \) Combining these, we get: \( I = -8 \log|x-2| + 11 \log|x-3| + c \) We can also write this as: \( I = 11 \log|x-3| - 8 \log|x-2| + c \) The constant of integration \( c \) is included in the final answer.
In simple words: When you have a fraction with different linear terms multiplied in the bottom, split it into simpler fractions using "partial fractions." Find the unknown constants by picking values for \( x \). Once separated, integrate each simple fraction, remembering that \( \int \frac{1}{ax+b} dx = \frac{1}{a} \log|ax+b| \).

🎯 Exam Tip: Partial fraction decomposition is crucial for integrating rational functions. Remember to set \( x \) to the roots of the denominator to quickly find the unknown constants \( A, B, C \).

Question 6. \( \frac { 4x^2+2x+6}{(x+1)^2(x-3)} \)
Answer:Let the given integral be \( I \). We need to integrate \( \frac{4x^2+2x+6}{(x+1)^2(x-3)} \) with respect to \( x \). This is a rational function with repeated linear factors and a distinct linear factor in the denominator. We use partial fraction decomposition. The form of the decomposition is: \( \frac{4x^2+2x+6}{(x+1)^2(x-3)} = \frac{A}{x-3} + \frac{B}{x+1} + \frac{C}{(x+1)^2} \) Multiply both sides by \( (x+1)^2(x-3) \): \( 4x^2+2x+6 = A(x+1)^2 + B(x-3)(x+1) + C(x-3) \) To find the constants \( A, B, C \): 1. **Set \( x=3 \) (to eliminate B and C):** \( 4(3)^2+2(3)+6 = A(3+1)^2 + B(0) + C(0) \) \( 4(9)+6+6 = A(4)^2 \) \( 36+12 = 16A \) \( 48 = 16A \implies A = \frac{48}{16} = 3 \) 2. **Set \( x=-1 \) (to eliminate A and B):** \( 4(-1)^2+2(-1)+6 = A(0) + B(0) + C(-1-3) \) \( 4(1)-2+6 = C(-4) \) \( 4-2+6 = -4C \) \( 8 = -4C \implies C = \frac{8}{-4} = -2 \) 3. **Set \( x=0 \) (a simple value, using A=3 and C=-2):** \( 4(0)^2+2(0)+6 = A(0+1)^2 + B(0-3)(0+1) + C(0-3) \) \( 6 = A(1)^2 + B(-3)(1) + C(-3) \) \( 6 = A - 3B - 3C \) Substitute \( A=3 \) and \( C=-2 \): \( 6 = 3 - 3B - 3(-2) \) \( 6 = 3 - 3B + 6 \) \( 6 = 9 - 3B \) \( 3B = 9 - 6 \) \( 3B = 3 \implies B = 1 \) So the partial fraction decomposition is: \( \frac{4x^2+2x+6}{(x+1)^2(x-3)} = \frac{3}{x-3} + \frac{1}{x+1} + \frac{-2}{(x+1)^2} \) Now, integrate each term: \( I = \int (\frac{3}{x-3} + \frac{1}{x+1} - \frac{2}{(x+1)^2}) \,dx \) \( \int \frac{3}{x-3} \,dx = 3 \log|x-3| \) \( \int \frac{1}{x+1} \,dx = \log|x+1| \) \( \int -\frac{2}{(x+1)^2} \,dx = -2 \int (x+1)^{-2} \,dx \) \( = -2 \frac{(x+1)^{-1}}{-1} = 2(x+1)^{-1} = \frac{2}{x+1} \) Combining these results: \( I = 3 \log|x-3| + \log|x+1| + \frac{2}{x+1} + c \) This is the final integral.
In simple words: First, break down the complex fraction into simpler parts using partial fractions. Since there's a squared term in the bottom, you'll have two fractions for it. Find the values for A, B, and C by plugging specific values for \( x \). Then, integrate each simple fraction separately. Remember that integrating \( 1/x \) gives \( \log|x| \), and integrating \( 1/x^2 \) gives \( -1/x \).

🎯 Exam Tip: When dealing with repeated linear factors like \( (ax+b)^n \) in partial fractions, remember to include terms for all powers from 1 to \( n \), i.e., \( \frac{B_1}{ax+b} + \frac{B_2}{(ax+b)^2} + \dots + \frac{B_n}{(ax+b)^n} \).

Question 7. \( \frac { 3x^2-2x+5}{(x-1)(x^2+5)} \)
Answer:Let the given integral be \( I \). We need to integrate \( \frac{3x^2-2x+5}{(x-1)(x^2+5)} \) with respect to \( x \). This rational function has a distinct linear factor and an irreducible quadratic factor in the denominator. We use partial fraction decomposition. The form of the decomposition is: \( \frac{3x^2-2x+5}{(x-1)(x^2+5)} = \frac{A}{x-1} + \frac{Bx+C}{x^2+5} \) Multiply both sides by \( (x-1)(x^2+5) \): \( 3x^2-2x+5 = A(x^2+5) + (Bx+C)(x-1) \) To find \( A, B, C \): 1. **Set \( x=1 \) (to eliminate the \( Bx+C \) term):** \( 3(1)^2-2(1)+5 = A(1^2+5) + (B(1)+C)(1-1) \) \( 3-2+5 = A(6) + 0 \) \( 6 = 6A \implies A = 1 \) 2. **Equate coefficients or choose other values for \( x \). Let's equate coefficients.** \( 3x^2-2x+5 = A(x^2+5) + Bx^2 - Bx + Cx - C \) \( 3x^2-2x+5 = Ax^2+5A + Bx^2 - Bx + Cx - C \) Group terms by powers of \( x \): \( 3x^2-2x+5 = (A+B)x^2 + (-B+C)x + (5A-C) \) Now, equate the coefficients: * **Coefficient of \( x^2 \):** \( A+B = 3 \) Since \( A=1 \), we have \( 1+B = 3 \implies B = 2 \) * **Coefficient of \( x \):** \( -B+C = -2 \) Since \( B=2 \), we have \( -2+C = -2 \implies C = 0 \) * **Constant term:** \( 5A-C = 5 \) Check with \( A=1, C=0 \): \( 5(1)-0 = 5 \). This is consistent. So the partial fraction decomposition is: \( \frac{3x^2-2x+5}{(x-1)(x^2+5)} = \frac{1}{x-1} + \frac{2x}{x^2+5} \) Now, integrate this expression: \( I = \int (\frac{1}{x-1} + \frac{2x}{x^2+5}) \,dx \) Integrate each term: \( \int \frac{1}{x-1} \,dx = \log|x-1| \) For the second term, \( \int \frac{2x}{x^2+5} \,dx \), notice that the numerator \( 2x \) is the derivative of the denominator \( x^2+5 \). So, \( \int \frac{f'(x)}{f(x)} \,dx = \log|f(x)| \). Therefore, \( \int \frac{2x}{x^2+5} \,dx = \log|x^2+5| \) (Since \( x^2+5 \) is always positive, we can write \( \log(x^2+5) \)). Combining these, we get: \( I = \log|x-1| + \log(x^2+5) + c \) Using logarithm properties, \( \log a + \log b = \log(ab) \): \( I = \log|(x-1)(x^2+5)| + c \) The source solution directly jumps to \( \log|x^3-x^2+5x-5| + c \). Note that \( (x-1)(x^2+5) = x^3+5x-x^2-5 = x^3-x^2+5x-5 \). So, the solution matches.
In simple words: For fractions with a quadratic term that cannot be factored in the bottom, use partial fractions with \( \frac{A}{linear} + \frac{Bx+C}{quadratic} \). Find A, B, C by setting \( x \) to roots or by matching coefficients. Integrate \( 1/(\text{linear}) \) to get \( \log \), and for \( \frac{f'(x)}{f(x)} \), the integral is also \( \log|f(x)| \). Combine the \( \log \) terms at the end.

🎯 Exam Tip: When integrating \( \frac{Bx+C}{Ax^2+Dx+E} \), always try to separate it into \( \frac{B}{2A} \frac{2Ax+D}{Ax^2+Dx+E} \) (which integrates to a log) and then a term like \( \frac{K}{Ax^2+Dx+E} \) (which often integrates to \( \tan^{-1} \) after completing the square). In this specific case, \( C=0 \), so it was simpler.

Question 8. Given f'(x) = \( \frac { 1 }{x} \) and f (1) = \( \frac {1}{π} \), then find f(x)
Answer:We are given the derivative of a function, \( f'(x) = \frac{1}{x} \), and a condition \( f(1) = \frac{1}{\pi} \). We need to find the original function \( f(x) \). To find \( f(x) \) from \( f'(x) \), we need to integrate \( f'(x) \) with respect to \( x \): \( f(x) = \int f'(x) \,dx \) \( f(x) = \int \frac{1}{x} \,dx \) The integral of \( \frac{1}{x} \) is \( \log|x| \). So, \( f(x) = \log|x| + c \) Here, \( c \) is the constant of integration. We use the given condition \( f(1) = \frac{1}{\pi} \) to find the value of \( c \). Substitute \( x=1 \) into the equation for \( f(x) \): \( f(1) = \log|1| + c \) We know that \( \log|1| \) (or \( \ln(1) \)) is equal to 0. So, \( f(1) = 0 + c \) We are given that \( f(1) = \frac{1}{\pi} \). Therefore, \( \frac{1}{\pi} = c \) Now, substitute the value of \( c \) back into the expression for \( f(x) \): \( f(x) = \log|x| + \frac{1}{\pi} \) This is the required function \( f(x) \). It's helpful to remember that \( \pi \) is a constant, so \( 1/\pi \) is also a constant.
In simple words: To find the original function from its derivative, you need to integrate it. The integral of \( 1/x \) is \( \log|x| \). This gives you \( f(x) \) plus a constant. Use the given extra information, \( f(1) = 1/\pi \), to find this constant. Since \( \log(1) \) is zero, the constant is \( 1/\pi \).

🎯 Exam Tip: When given a derivative and a point on the original function, always integrate first to find the general form with a constant of integration (\( c \)), then use the given point to solve for that specific constant.