OP Malhotra Class 12 Maths Solutions Chapter 13 Indefinite Integral 1 Exercise 13 (B)

Question 1. \( \int \frac{\left(x^2-1\right)^2}{x^3} d x \)
Answer: First, we expand the square in the numerator and then divide each term by \( x^3 \).
\[ \int \frac{\left(x^2-1\right)^2}{x^3} d x = \int \frac{x^4+1-2x^2}{x^3} d x \]
\( \implies \) Next, we split the fraction into simpler terms to integrate them individually.
\[ = \int\left(x+\frac{1}{x^3}-\frac{2}{x}\right)d x \]
\( \implies \) Now, we apply the power rule for integration (\( \int x^n dx = \frac{x^{n+1}}{n+1} \)) and the logarithm rule (\( \int \frac{1}{x} dx = \log|x| \)).
\[ = \frac{x^2}{2} + \frac{x^{-3+1}}{-3+1} - 2\log|x| + C \]
\( \implies \) Simplify the terms to get the final answer.
\[ = \frac{x^2}{2} - \frac{1}{2x^2} - 2\log|x| + C \]In simple words: First, open up the bracket on top, then divide every part by \( x^3 \). After that, integrate each simple piece separately. Remember that integrating \( 1/x \) gives you \( \log|x| \).

🎯 Exam Tip: Always expand algebraic expressions and simplify them by dividing before integrating, as this often converts complex fractions into easily integrable power terms and logarithmic forms.

Question 2. \( \int \frac{x+1}{x^2-1} d x \)
Answer: To solve this integral, we first simplify the fraction by factoring the denominator.
\[ \int \frac{x+1}{x^2-1} d x = \int \frac{(x+1)}{(x-1)(x+1)} d x \]
\( \implies \) We can cancel out the common term \( (x+1) \) from the numerator and denominator.
\[ = \int \frac{d x}{x-1} \]
\( \implies \) This is a standard integral form, where the integral of \( \frac{1}{ax+b} \) is \( \frac{1}{a} \log|ax+b| \). Here, \( a=1 \).
\[ = \log|x-1| + C \]In simple words: First, break down the bottom part of the fraction using \( a^2-b^2 = (a-b)(a+b) \). Then, cancel out matching parts from the top and bottom. What's left is a simple integral that turns into a logarithm.

🎯 Exam Tip: Always look for opportunities to factorize and simplify the integrand before integration, as it can often reduce a complex problem to a basic standard form.

Question 3. \( \int e^{-x} d x \)
Answer: We use the standard integration formula for exponential functions: \( \int e^{mx} dx = \frac{e^{mx}}{m} + C \).
\[ \int e^{-x} d x = \frac{e^{-x}}{-1} + C \]
\( \implies \) Simplify the expression.
\[ = -e^{-x} + C \] This formula is very helpful for quick exponential integration.
In simple words: When you integrate \( e \) to the power of something like \( -x \), you just write \( e \) to that same power, and then divide it by the number in front of the \( x \) (which is \( -1 \) here).

🎯 Exam Tip: Remember that for \( \int e^{kx} dx \), you divide by the coefficient of \( x \), not multiply. Pay close attention to negative signs in the exponent.

Question 4. \( \int e^{3 x} d x \)
Answer: We apply the general formula for integrating exponential functions, which is \( \int e^{mx} dx = \frac{e^{mx}}{m} + C \).
\[ \int e^{3 x} d x = \frac{e^{3 x}}{3} + C \] The constant \( C \) is the integration constant.
In simple words: To integrate \( e \) raised to the power of \( 3x \), you simply write \( e \) to the power of \( 3x \) and then divide it by the number \( 3 \). Don't forget to add \( C \).

🎯 Exam Tip: For integrals of the form \( \int e^{kx} dx \), the result is always \( \frac{1}{k}e^{kx} + C \). Make sure to divide by the constant \( k \).

Question 5. \( \int a^{2 x} dx \)
Answer: We use the standard integration formula for exponential functions with a base other than \( e \): \( \int a^{mx} dx = \frac{a^{mx}}{m \log a} + C \).
\[ \int a^{2 x} dx = \frac{a^{2x}}{2(\log a)} + C \] This formula is an extension of the basic exponential integral rule.
In simple words: When you integrate a number \( a \) raised to the power of \( 2x \), you write \( a \) to the power of \( 2x \), then divide by \( 2 \) times the natural logarithm of \( a \).

🎯 Exam Tip: Be careful with the base \( a \). The formula for \( \int a^{kx} dx \) requires \( \log a \) in the denominator, in addition to the coefficient \( k \).

Question 6. \( \int\left(e^{3 a \log _e x}+e^{3 x \log _e a}\right) d x \)
Answer: Let \( I \) be the given integral. We use logarithm properties first: \( k \log y = \log y^k \) and \( e^{\log_e y} = y \).
\[ I = \int\left(e^{3 a \log _e x}+e^{3 x \log _e a}\right) d x \]
\( \implies \) Apply the logarithm property \( k \log y = \log y^k \).
\[ = \int\left(e^{\log _e x^{3 a}}+e^{\log _e a^{3 x}}\right) d x \]
\( \implies \) Apply the property \( e^{\log_e y} = y \).
\[ = \int x^{3 a} d x+\int a^{3 x} d x \]
\( \implies \) Now, integrate each term using the power rule (\( \int x^n dx = \frac{x^{n+1}}{n+1} \)) and the exponential rule (\( \int a^{mx} dx = \frac{a^{mx}}{m \log a} \)).
\[ = \frac{x^{3 a+1}}{3 a+1} + \frac{a^{3 x}}{3 \log a} + C \]In simple words: First, use log rules to simplify the powers of \( e \). Remember that \( e \) raised to the power of \( \log x \) is just \( x \). After simplifying, you will have two simple integrals: one for \( x \) to a power and one for \( a \) to a power. Integrate them separately.

🎯 Exam Tip: Always simplify expressions involving logarithms and exponentials using their properties before attempting integration. Look for `e^(log x)` patterns to simplify to `x`.

Question 7. \( \int \frac{3 e^{2 x}+3 e^{4 x}}{e^x+e^{-x}} d x \)
Answer: To integrate this, we need to simplify the fraction first. Multiply the numerator and denominator by \( e^x \).
\[ \int \frac{3 e^{2 x}+3 e^{4 x}}{e^x+e^{-x}} d x \]
\( \implies \) Factor out \( 3e^{2x} \) from the numerator and rearrange the denominator.
\[ =\int \frac{3 e^{2x}(1+e^{2 x})}{(e^{2 x}+1) e^{-x}} d x \]
\( \implies \) Note the original denominator \( e^x+e^{-x} \). Multiplying numerator and denominator by \( e^x \) would give: Numerator: \( (3 e^{2x}+3 e^{4x})e^x = 3e^{3x}+3e^{5x} \) Denominator: \( (e^x+e^{-x})e^x = e^{2x}+e^0 = e^{2x}+1 \). So the expression becomes: \[ \int \frac{3e^{3x}(1+e^{2x})}{e^{2x}+1} dx \]
\( \implies \) Cancel the common term \( (1+e^{2x}) \).
\[ =\int 3 e^{3 x} d x \]
\( \implies \) Now, integrate using the formula \( \int e^{mx} dx = \frac{e^{mx}}{m} + C \).
\[ =3 \frac{e^{3 x}}{3}+C \]
\( \implies \) Simplify the result.
\[ =e^{3 x}+C \]In simple words: First, rewrite the bottom of the fraction to make it easier to work with. Then, take out common parts from the top and bottom to make the fraction simpler. Once it's simple, integrate it like a normal \( e \) to the power of something.

🎯 Exam Tip: When dealing with exponential fractions, try multiplying the numerator and denominator by `e^x` or `e^-x` to simplify the expression and eliminate negative exponents, making it easier to integrate.

Question 8. \( \int\left(\frac{5 x+7}{x}+e^x\right) d x \)
Answer: We can split the fraction into individual terms before integrating.
\[ \int\left(\frac{5 x+7}{x}+e^x\right) d x = \int\left[5+\frac{7}{x}+e^x\right] d x \]
\( \implies \) Now, integrate each term separately using standard formulas: \( \int k dx = kx \), \( \int \frac{1}{x} dx = \log|x| \), and \( \int e^x dx = e^x \).
\[ =\int 5 d x+7 \int \frac{1}{x} d x+\int e^x d x \]
\( \implies \) Apply the integration rules.
\[ =5 x+7 \log |x|+e^x+C \]In simple words: Break the fraction into separate parts, \( 5x/x \) and \( 7/x \). Then integrate each part one by one. Remember that \( 1/x \) becomes \( \log|x| \) and \( e^x \) stays \( e^x \).

🎯 Exam Tip: Always split fractions where the numerator is a sum/difference over a single term in the denominator. This allows term-by-term integration using basic formulas.

Question 9. \( \int \frac{a x^2+bx+c}{x^2} d x \)
Answer: Let \( I \) be the given integral. We start by splitting the fraction into simpler terms.
\[ I = \int \frac{a x^2+bx+c}{x^2} d x \]
\( \implies \) Divide each term in the numerator by \( x^2 \).
\[ = \int\left[a+\frac{b}{x}+\frac{c}{x^2}\right] d x \]
\( \implies \) Now, integrate each term separately. Remember that \( \int x^n dx = \frac{x^{n+1}}{n+1} \) and \( \int \frac{1}{x} dx = \log|x| \).
\[ = a x+b \log |x|-\frac{c}{x}+C' \] The constant \( C' \) is the integration constant.
In simple words: First, divide each part of the top by \( x^2 \). This makes three simple parts. Then, integrate each part separately. Remember that \( c/x^2 \) can be written as \( c x^{-2} \) for easier integration.

🎯 Exam Tip: When the numerator has multiple terms and the denominator is a single term, divide each term in the numerator by the denominator to simplify before integrating.

Question 10. \( \int \frac{d x}{\sqrt{16-x^2}} \)
Answer: This integral is in a standard form for an inverse trigonometric function. We recognize it as \( \int \frac{d x}{\sqrt{a^2-x^2}} = \sin^{-1} \frac{x}{a} + C \).
\[ \int \frac{d x}{\sqrt{16-x^2}} = \int \frac{d x}{\sqrt{4^2-x^2}} \]
\( \implies \) Here, \( a=4 \). Apply the formula directly.
\[ = \sin^{-1} \frac{x}{4}+C \] This is a direct application of the formula for inverse sine. \[ \therefore \int \frac{d x}{\sqrt{a^2-x^2}} = \sin^{-1} \frac{x}{a} + C \]In simple words: This problem matches a special rule for integration. We see that \( 16 \) is \( 4^2 \). So, the answer is simply \( \sin^{-1} \) of \( x \) divided by \( 4 \).

🎯 Exam Tip: Memorize the standard inverse trigonometric integral formulas. Recognizing `a^2-x^2` in the denominator with a square root immediately points to `sin^-1`.

Question 11. \( \int \frac{d x}{\sqrt{25-4 x^2}} \)
Answer: To use the standard \( \sin^{-1} \) formula, the coefficient of \( x^2 \) must be 1. So, we factor out 4 from the term under the square root.
\[ \int \frac{d x}{\sqrt{25-4 x^2}} = \int \frac{d x}{\sqrt{4(\frac{25}{4}-x^2)}} \]
\( \implies \) Take \( \sqrt{4} = 2 \) out of the square root.
\[ = \frac{1}{2} \int \frac{d x}{\sqrt{\left(\frac{5}{2}\right)^2-x^2}} \]
\( \implies \) Now this is in the form \( \int \frac{d x}{\sqrt{a^2-x^2}} \) where \( a=\frac{5}{2} \). Apply the formula \( \sin^{-1} \frac{x}{a} \).
\[ = \frac{1}{2}\sin^{-1}\frac{x}{(5/2)}+C \]
\( \implies \) Simplify the fraction inside \( \sin^{-1} \).
\[ = \frac{1}{2}\sin^{-1} \frac{2x}{5}+C \] This method helps convert the integral into a known form.
\[ \therefore \int \frac{d x}{\sqrt{a^2-x^2}} = \sin^{-1} \frac{x}{a} + C \]In simple words: First, change the bottom part so that \( x^2 \) doesn't have a number in front of it. You do this by taking the number \( 4 \) out from under the square root, which becomes \( 2 \) outside. Then, use the special rule for \( \sin^{-1} \) by putting \( x \) over the new 'a' value.

🎯 Exam Tip: When the `x^2` term has a coefficient other than 1, always factor out that coefficient from under the square root to match the standard `sin^-1` integral form correctly.

Question 12. \( \int \frac{d x}{4+x^2} \)
Answer: This integral matches the standard form for inverse tangent: \( \int \frac{d x}{a^2+x^2} = \frac{1}{a} \tan^{-1} \frac{x}{a} + C \).
\[ \int \frac{d x}{4+x^2} = \int \frac{d x}{2^2+x^2} \]
\( \implies \) Here, \( a=2 \). Apply the formula directly.
\[ = \frac{1}{2} \tan^{-1}\left(\frac{x}{2}\right)+C \] This formula is very important for solving integrals of this type. \[ \therefore \int \frac{d x}{a^2+x^2} = \frac{1}{a} \tan^{-1} \frac{x}{a} + C \]In simple words: This integral fits a known pattern for the inverse tangent function. Since \( 4 \) is \( 2^2 \), our 'a' value is \( 2 \). So, the answer is \( 1/2 \) times \( \tan^{-1} \) of \( x \) divided by \( 2 \).

🎯 Exam Tip: Memorize the standard inverse tangent integral formula. Recognize `a^2+x^2` in the denominator without a square root immediately points to `tan^-1`.

Question 13. \( \int \frac{d x}{16+9 x^2} \)
Answer: To use the standard \( \tan^{-1} \) formula, the coefficient of \( x^2 \) must be 1. We factor out 9 from the denominator.
\[ \int \frac{d x}{16+9 x^2} = \int \frac{d x}{9\left(\frac{16}{9}+x^2\right)} \]
\( \implies \) Take \( \frac{1}{9} \) outside the integral.
\[ = \frac{1}{9} \int \frac{d x}{\left(\frac{4}{3}\right)^2+x^2} \]
\( \implies \) Now this is in the form \( \int \frac{d x}{a^2+x^2} \) where \( a=\frac{4}{3} \). Apply the formula \( \frac{1}{a} \tan^{-1} \frac{x}{a} \).
\[ = \frac{1}{9} \cdot \frac{1}{(4/3)} \tan^{-1}\frac{x}{(4/3)}+C \]
\( \implies \) Simplify the constants and the fraction inside \( \tan^{-1} \).
\[ = \frac{1}{9} \cdot \frac{3}{4} \tan^{-1}\frac{3x}{4}+C \]
\( \implies \) Multiply the fractions.
\[ = \frac{1}{12} \tan^{-1}\left(\frac{3x}{4}\right)+C \] This process helps transform the integral into a recognizable form.
In simple words: Make sure the \( x^2 \) part at the bottom has no number in front of it. Take the \( 9 \) out of the denominator. Then, find the 'a' value (which is \( 4/3 \)) and use the special rule for \( \tan^{-1} \).

🎯 Exam Tip: For integrals of the form `1/(c + dx^2)`, always factor out `d` from the denominator to make the coefficient of `x^2` equal to 1. This prepares the integral for the standard `tan^-1` formula.

Question 14. \( \int \left(\frac{6}{1+x^2}+10^x-5 \operatorname{cosec}^2 x\right) d x \)
Answer: We can integrate each term separately as the integral of a sum/difference is the sum/difference of the integrals.
\[ \text{Sol. } = \int \frac{6}{1+x^2} d x+\int 10^x d x - \int 5 \operatorname{cosec}^2 x d x \]
\( \implies \) Take constants outside the integral sign.
\[ = 6 \int \frac{d x}{1+x^2}+ \int 10^x d x - 5 \int \operatorname{cosec}^2 x d x \]
\( \implies \) Apply standard integration formulas: \( \int \frac{d x}{1+x^2} = \tan^{-1} x \) \( \int a^x dx = \frac{a^x}{\log a} \) \( \int \operatorname{cosec}^2 x d x = -\cot x \).
\[ = 6 \tan^{-1}x + \frac{10^x}{\log 10} - 5 (-\cot x) + C \]
\( \implies \) Simplify the last term.
\[ = 6 \tan^{-1}x + \frac{10^x}{\log 10} + 5 \cot x + C \] Each term is integrated using its specific rule.
In simple words: Break the big problem into three smaller problems, one for each part. Integrate \( 6/(1+x^2) \) to get \( 6 \tan^{-1}x \). Integrate \( 10^x \) using its special rule. Integrate \( -5 \operatorname{cosec}^2 x \) using its rule, remembering that \( \operatorname{cosec}^2 x \) gives \( -\cot x \).

🎯 Exam Tip: For integrals of sums or differences of functions, integrate each term separately. Make sure to recall and apply the correct standard integral formula for each function type (inverse tan, exponential, trigonometric).

Question 15. \( \int\left(x+\frac{1}{x}\right)\left(x^2+\frac{1}{x^2}\right) d x \)
Answer: First, we need to expand the product of the two terms in the integrand.
\[ \int\left(x+\frac{1}{x}\right)\left(x^2+\frac{1}{x^2}\right) d x \]
\( \implies \) Multiply the terms: \( x \cdot x^2 = x^3 \), \( x \cdot \frac{1}{x^2} = \frac{1}{x} \), \( \frac{1}{x} \cdot x^2 = x \), \( \frac{1}{x} \cdot \frac{1}{x^2} = \frac{1}{x^3} \).
\[ =\int\left(x^3+\frac{1}{x}+x+\frac{1}{x^3}\right) d x \]
\( \implies \) Now, integrate each term using the power rule (\( \int x^n dx = \frac{x^{n+1}}{n+1} \)) and \( \int \frac{1}{x} dx = \log|x| \).
\[ =\frac{x^4}{4}+\log |x|+\frac{x^2}{2}+\frac{x^{-3+1}}{-3+1}+C \]
\( \implies \) Simplify the last term.
\[ =\frac{x^4}{4}+\log |x|+\frac{x^2}{2}-\frac{1}{2 x^2}+C \] Expanding the product first simplifies the problem greatly.
In simple words: Multiply the two brackets together, just like you multiply numbers. This will give you four separate parts. Then, integrate each part by itself. Remember that \( 1/x \) becomes \( \log|x| \) and \( 1/x^3 \) becomes \( x^{-3} \) which you integrate using the power rule.

🎯 Exam Tip: Always simplify and expand products within an integrand algebraically before integrating. This transforms the expression into a sum of power functions, which are easier to integrate.

Question 16. \( \int \frac{(x+2)\left(4 x^2-5\right)}{x} d x \)
Answer: First, expand the numerator by multiplying the two factors, then divide each term by \( x \).
\[ \int \frac{(x+2)\left(4 x^2-5\right)}{x} d x \]
\( \implies \) Expand the numerator: \( (x+2)(4x^2-5) = 4x^3 - 5x + 8x^2 - 10 \).
\[ =\int \frac{4 x^3+8 x^2-5 x-10}{x} d x \]
\( \implies \) Divide each term in the numerator by \( x \).
\[ =4 \int x^2 d x+8 \int x d x-\int 5 d x-\int \frac{10}{x} d x \]
\( \implies \) Now, integrate each term using the power rule (\( \int x^n dx = \frac{x^{n+1}}{n+1} \)) and \( \int \frac{1}{x} dx = \log|x| \).
\[ =4 \frac{x^3}{3}+8 \frac{x^2}{2}-5 x-10 \log |x|+C \]
\( \implies \) Simplify the coefficients.
\[ =4 \frac{x^3}{3}+4 x^2-5 x-10 \log |x|+C \] This method helps convert a complex fraction into a sum of basic power functions and logarithms.
In simple words: First, multiply the two parts on top together. Then, divide every part of that new top expression by \( x \). This will give you a list of simple terms to integrate one by one.

🎯 Exam Tip: When faced with a product in the numerator and a single term in the denominator, always expand the product first, then divide each term by the denominator before integrating.

Question 17. \( \int \frac{x^2}{4+x^2} d x \)
Answer: Let \( I \) be the given integral. When the degree of the numerator is equal to or greater than the degree of the denominator, we often use algebraic manipulation or polynomial division.
\[ I=\int \frac{x^2}{4+x^2} d x \]
\( \implies \) Add and subtract 4 in the numerator to create a term that matches the denominator.
\[ =\int \frac{4+x^2-4}{4+x^2} d x=\int\left[1-\frac{4}{x^2+4}\right] d x \]
\( \implies \) Separate the terms and integrate. The first term is a constant, and the second is in the \( \tan^{-1} \) form.
\[ =x-4 \int \frac{d x}{x^2+2^2} \]
\( \implies \) Apply the formula \( \int \frac{d x}{a^2+x^2} = \frac{1}{a} \tan^{-1} \frac{x}{a} + C \), where \( a=2 \).
\[ =x-\frac{4}{2} \tan ^{-1} \frac{x}{2}+C \]
\( \implies \) Simplify the coefficient.
\[ =x-2 \tan ^{-1} \frac{x}{2}+C \] This technique is useful for such rational functions.
In simple words: When the top and bottom of the fraction have the same highest power, try to make the top look like the bottom. Here, we add and subtract \( 4 \) in the top. This splits the fraction into a simple \( 1 \) and another part that fits the \( \tan^{-1} \) rule.

🎯 Exam Tip: For rational functions where the degree of the numerator is equal to or greater than the degree of the denominator, perform algebraic manipulation (like adding and subtracting terms) or polynomial long division to simplify the integrand before integrating.

Question 18. \( \int \frac{x^4}{1+x^2} d x \)
Answer: Let \( I \) be the given integral. We use a similar algebraic manipulation technique as in Question 17.
\[ I = \int \frac{x^4}{1+x^2} d x = \int \frac{x^4-1+1}{1+x^2} d x \]
\( \implies \) Split the fraction into two parts.
\[ =\int \frac{x^4-1}{1+x^2} d x+\int \frac{d x}{1+x^2} \]
\( \implies \) Factorize \( x^4-1 \) as \( (x^2-1)(x^2+1) \).
\[ =\int \frac{\left(x^2-1\right)\left(x^2+1\right)}{1+x^2} d x+\int \frac{d x}{1+x^2} \]
\( \implies \) Cancel the common term \( (x^2+1) \).
\[ =\int \left(x^2-1\right) d x+\int \frac{d x}{1+x^2} \]
\( \implies \) Integrate each term. The first integral uses the power rule, and the second is a standard \( \tan^{-1} \) form.
\[ =\frac{x^3}{3}-x+\tan ^{-1} x+c \] This approach simplifies the integral significantly.
In simple words: Add and subtract \( 1 \) from the top, then split the fraction. Use the rule \( a^2-b^2 = (a-b)(a+b) \) to break \( x^4-1 \) into \( (x^2-1)(x^2+1) \). This lets you cancel parts, leaving simple terms and a \( \tan^{-1} \) integral.

🎯 Exam Tip: For integrals involving polynomials, look for opportunities to add and subtract terms in the numerator to simplify the fraction by matching part of the denominator, especially when factorization identities like `a^2-b^2` can be used.

Question 19. \( \int \frac{x^6-1}{x^2+1} d x \)
Answer: Let \( I \) be the given integral. We use algebraic manipulation and factorization for this problem.
\[ I=\int \frac{x^6-1}{x^2+1} d x \]
\( \implies \) Rewrite \( x^6-1 \) as \( (x^2)^3 - 1^3 \). Using the identity \( a^3-b^3 = (a-b)(a^2+ab+b^2) \), where \( a=x^2 \) and \( b=1 \).
\[ = \int \frac{(x^2-1)(x^4+x^2+1)}{x^2+1} d x \]
*Correction in source, following standard algebraic method:* Alternatively, we can use polynomial long division or algebraic adjustment: \[ = \int \frac{x^6+1-2}{x^2+1} d x \]
\( \implies \) Split the integral into two parts.
\[ = \int \frac{x^6+1}{x^2+1} d x - \int \frac{2}{x^2+1} d x \]
\( \implies \) Use the factorization \( a^3+b^3 = (a+b)(a^2-ab+b^2) \), where \( a=x^2 \) and \( b=1 \). So, \( x^6+1 = (x^2)^3+1^3 = (x^2+1)(x^4-x^2+1) \).
\[ = \int \frac{(x^2+1)(x^4-x^2+1)}{x^2+1} d x - \int \frac{2}{x^2+1} d x \]
\( \implies \) Cancel the common term \( (x^2+1) \).
\[ = \int (x^4-x^2+1) d x - 2 \int \frac{1}{x^2+1} d x \]
\( \implies \) Integrate each term using the power rule and the \( \tan^{-1} \) formula.
\[ = \frac{x^5}{5}-\frac{x^3}{3}+x - 2 \tan ^{-1} x+C \] This method provides a clear path to the solution by using algebraic identities.
In simple words: Think of \( x^6-1 \) as \( (x^2)^3-1^3 \). Use the rule for \( a^3-b^3 \) to break it down. Then, divide by \( x^2+1 \) to simplify the fraction. After that, you'll have simple power terms and a \( \tan^{-1} \) integral to solve.

🎯 Exam Tip: When integrating rational functions involving higher powers like `x^6`, try to use algebraic identities for sums or differences of cubes (`a^3 ± b^3`) or powers (`a^n ± b^n`) to factorize the numerator and simplify the expression with the denominator.

Question 20. \( \int \frac{e^{6 \log x}-e^{5 \log x}}{e^{5 \log x}-e^{3 \log x}} d x \)
Answer: Let \( I \) be the given integral. First, simplify the exponential-logarithmic terms using the properties \( k \log y = \log y^k \) and \( e^{\log y} = y \).
\[ I = \int \frac{e^{6 \log x}-e^{5 \log x}}{e^{5 \log x}-e^{3 \log x}} d x \]
\( \implies \) Apply \( k \log x = \log x^k \).
\[ =\int \frac{e^{\log x^6}-e^{\log x^5}}{e^{\log x^5}-e^{\log x^3}} d x \]
\( \implies \) Apply \( e^{\log x} = x \).
\[ =\int \frac{x^6-x^5}{x^5-x^3} d x \]
\( \implies \) Factor out common terms from the numerator (\( x^5 \)) and the denominator (\( x^3 \)).
\[ =\int \frac{x^5(x-1)}{x^3\left(x^2-1\right)} d x \]
\( \implies \) Cancel \( x^3 \) and factorize \( (x^2-1) \) as \( (x-1)(x+1) \).
\[ =\int \frac{x^2(x-1)}{(x-1)(x+1)} d x \]
\( \implies \) Cancel \( (x-1) \).
\[ =\int \frac{x^2}{x+1} d x \]
\( \implies \) Now, use algebraic manipulation similar to Question 17: add and subtract 1 in the numerator.
\[ =\int \frac{x^2-1+1}{x+1} d x=\int\left[(x-1)+\frac{1}{x+1}\right] d x \]
\( \implies \) Integrate each term separately.
\[ =\frac{x^2}{2}-x+\log (x+1)+C \] This solution demonstrates the power of simplification before integration.
In simple words: First, use log rules to change `e^(k log x)` into `x^k`. This makes the whole fraction simpler with just powers of \( x \). Then, take out common parts from the top and bottom to simplify the fraction even more. Finally, split the last fraction into easy parts and integrate each part.

🎯 Exam Tip: When an integrand involves `e` raised to logarithmic powers, always simplify using logarithm properties first (`k log x = log x^k` and `e^(log x) = x`). Then, use algebraic factorization and simplification techniques before integrating.