Maharashtra Board Class 12 Maths Part 1 Chapter 5 Integration 5.2 Solutions

Evaluate The Following.

Question 1. \( \int x\sqrt{1 + x^2}dx \)
Answer: Solution: Let \( I = \int x\sqrt{1 + x^2} dx \) Put \( 1 + x^2 = t \)
\( \implies 2xdx = dt \)
\( \implies xdx = \frac{dt}{2} \)
\( \implies I = \int \sqrt{t} \frac{dt}{2} = \frac{1}{2} \int t^{1/2}dt \) \( = \frac{1}{2} \frac{t^{3/2}}{3/2} + c \) \( = \frac{1}{3} (1 + x^2)^{3/2} + c. \)
In simple words: This problem is solved using the substitution method by letting \( t = 1 + x^2 \). Its derivative \( 2x dx \) simplifies the integrand into a basic power rule integral. After integrating with respect to \( t \), substitute back \( x^2 \) for \( t \) to get the final answer.

🎯 Exam Tip: When using substitution, always ensure all parts of the original integral involving \( x \) are converted to \( t \) before integration. Don't forget to add the constant of integration, `+ c`, and re-substitute the original variable.

Question 2. \( \int \frac{x^3}{\sqrt{1+x^4}}dx \)
Answer: Solution: Let \( I = \int \frac{x^3}{\sqrt{1+x^4}}dx \) Put \( 1 + x^4 = t \)
\( \implies 4x^3dx = dt \)
\( \implies x^3dx = \frac{dt}{4} \)
\( \implies I = \int \frac{1}{\sqrt{t}} \cdot \frac{dt}{4} = \frac{1}{4} \int t^{-1/2}dt \) \( = \frac{1}{4} \frac{t^{1/2}}{1/2} + c \) \( = \frac{1}{2} \sqrt{1 + x^4} + c. \)
In simple words: This integral uses the substitution method by setting \( t = 1 + x^4 \). The derivative \( 4x^3 dx \) matches the \( x^3 dx \) term in the numerator, simplifying the integral to a basic power form for integration.

🎯 Exam Tip: Pay close attention to the powers of \( x \) in the integrand. Often, if you see an \( x^n \) term and an \( x^{n-1} \) term, a substitution of the \( x^n \) part (or a related expression) is effective. Always double-check the constant factor derived from the substitution.

Question 3. \( \int (e^x + e^{-x})^2 (e^x – e^{-x})dx \)
Answer: Solution: Let \( I = \int (e^x + e^{-x})^2 (e^x - e^{-x})dx \) Put \( e^x + e^{-x} = t \)
\( \implies (e^x - e^{-x}) dx = dt \)
\( \implies I = \int t^2dt = \frac{t^3}{3} + c \) \( = \frac{(e^x + e^{-x})^3}{3} + c. \)
In simple words: This integral is solved by observing that the derivative of the base of the squared term, \( (e^x + e^{-x}) \), is exactly \( (e^x - e^{-x}) \). This allows for a direct substitution, transforming it into a simple power rule integral.

🎯 Exam Tip: Recognize common derivative pairs. The derivative of \( e^x + e^{-x} \) is \( e^x - e^{-x} \), and the derivative of \( e^x - e^{-x} \) is \( e^x + e^{-x} \). This pattern frequently indicates a direct substitution. Be careful with signs.

Question 4. \( \int \frac{1+x}{x+e^{-x}}dx \)
Answer: Solution: Let \( I = \int \frac{1+x}{x+e^{-x}} dx \) \( = \int \frac{(1+x)e^x}{(x+e^{-x})e^x} dx \) \( = \int \frac{e^x+xe^x}{xe^x+1} dx \) Put \( xe^x + 1 = t \)
\( \implies (x \cdot e^x + e^x \cdot 1) dx = dt \)
\( \implies (1+x)e^x dx = dt \)
\( \implies I = \int \frac{1}{t} dt = \log |t| + c \) \( = \log |xe^x + 1| + c. \)
In simple words: To integrate this, we multiply the numerator and denominator by \( e^x \) to transform the expression. This makes the denominator \( xe^x + 1 \), whose derivative \( (1+x)e^x \) is conveniently found in the numerator, allowing for a logarithmic integral after substitution.

🎯 Exam Tip: When the denominator involves a sum with an exponential term, multiplying by \( e^x \) or \( e^{-x} \) can sometimes simplify the expression and reveal a suitable substitution. Look for derivatives that match the numerator after such a manipulation.

Question 5. \( \int (x + 1)(x + 2)(x + 3) dx \)
Answer: Solution: Let \( I = \int (x + 1)(x + 2)^7(x + 3) dx \) \( = \int (x + 2)^7 (x + 1)(x + 3) dx \) \( = \int (x + 2)^7 [(x + 2) – 1][(x + 2) + 1] dx \) \( = \int (x + 2)^7 [(x + 2)^2 – 1] dx \) \( = \int [(x + 2)^9 – (x + 2)^7] dx \) \( = \int (x + 2)^9 dx - \int (x + 2)^7 dx \) \( = \frac{(x+2)^{10}}{10} - \frac{(x+2)^8}{8} + C \)
In simple words: This integral is simplified by making a substitution \( u = x+2 \). We then express \( x+1 \) and \( x+3 \) in terms of \( u \) to create a polynomial in \( u \), which can be integrated term by term using the power rule.

🎯 Exam Tip: If you have multiple linear factors like \( (x+a) \), \( (x+b) \), etc., try to make a substitution that simplifies the most complex or frequently appearing term, like \( x+2 \) here. This often reduces the problem to a standard polynomial integration.

Question 6. \( \int \frac{1}{x \log x} dx \)
Answer: Solution: Put \( \log x = t \)
\( \implies \frac{1}{x} dx = dt \)
\( \implies I = \int \frac{1}{\log x} \cdot \frac{1}{x} dx \) \( = \int \frac{1}{t} dt \) \( = \log |t| + c \) \( = \log |\log x| + c. \)
In simple words: This integral is solved by substituting \( t = \log x \). The derivative of \( \log x \), which is \( \frac{1}{x} \), is present in the integrand, converting the problem into a basic \( \int \frac{1}{t} dt \) form, whose integral is \( \log|t| \).

🎯 Exam Tip: The derivative of \( \log x \) is \( \frac{1}{x} \). Whenever you see \( \log x \) multiplied by \( \frac{1}{x} \) (or divided by \( x \)) in an integrand, consider substituting \( t = \log x \) for a quick solution.

Question 7. \( \int \frac{x^5}{x^2+1} dx \)
Answer: Solution: Let \( I = \int \frac{x^5}{x^2+1} dx \) \( = \int \frac{x^3 \cdot x^2}{x^2+1} dx \) Put \( x^2 + 1 = t \)
\( \implies 2x dx = dt \)
\( \implies x dx = \frac{dt}{2} \) and \( x^2 = t-1 \)
\( \implies I = \int \frac{(t-1) \cdot x^2}{t} \frac{dt}{2x} \) (Incorrect step in OCR, it should be: \( I = \int \frac{(t-1) \cdot (t-1)}{t} \frac{dt}{2} \) if we had \( x^4 \cdot x dx \). The original OCR skips steps here.) Let's carefully re-trace the OCR steps for Q7. \( I = \int \frac{x^5}{x^2+1} dx \) \( = \int \frac{x^4 \cdot x}{x^2+1} dx \) (This is a missing intermediate step from the OCR solution) Put \( x^2+1 = t \)
\( \implies 2x dx = dt \)
\( \implies x dx = \frac{dt}{2} \) and \( x^2 = t-1 \), so \( x^4 = (t-1)^2 \)
\( \implies I = \int \frac{(t-1)^2}{t} \frac{dt}{2} \) \( = \frac{1}{2} \int \frac{t^2-2t+1}{t} dt \) \( = \frac{1}{2} \int (t-2+\frac{1}{t}) dt \) \( = \frac{1}{2} \left[ \int t dt - \int 2 dt + \int \frac{1}{t} dt \right] \) \( = \frac{1}{2} \left[ \frac{t^2}{2} - 2t + \log |t| \right] + c \) \( = \frac{1}{2} \left[ \frac{(x^2+1)^2}{2} - 2(x^2+1) + \log |x^2+1| \right] + c. \)
In simple words: This integral requires a substitution \( t = x^2+1 \). We then express \( x^5 \) as \( x^4 \cdot x \) and substitute \( x^4 = (t-1)^2 \) and \( x dx = \frac{dt}{2} \). The resulting polynomial in \( t \) is then integrated term by term.

🎯 Exam Tip: For rational functions where the degree of the numerator is greater than or equal to the denominator, long division or algebraic manipulation (like adding and subtracting terms) can simplify the integrand. In substitution, ensure all powers of \( x \) are converted to \( t \) and the differential \( dx \) is correctly replaced by \( dt \).

Question 8. \( \int \frac{2x+6}{\sqrt{x^2+6x+3}} dx \)
Answer: Solution: Let \( I = \int \frac{2x+6}{\sqrt{x^2+6x+3}} dx \) Put \( x^2 + 6x + 3 = t \)
\( \implies (2x+6)dx = dt \)
\( \implies I = \int \frac{1}{\sqrt{t}} dt = \int t^{-1/2}dt \) \( = \frac{t^{1/2}}{1/2} + c \) \( = 2\sqrt{x^2 + 6x + 3} + c. \)
In simple words: This integral is straightforward using substitution. By letting \( t = x^2+6x+3 \), its derivative \( (2x+6)dx \) directly matches the numerator, simplifying the integral to a basic power form \( \int t^{-1/2} dt \).

🎯 Exam Tip: Always look for a function whose derivative is present in the integrand. If the numerator is the derivative of the expression inside a square root in the denominator, a simple substitution often yields a quick solution. Remember \( \int \frac{f'(x)}{\sqrt{f(x)}} dx = 2\sqrt{f(x)} + c \).

Question 9. \( \int \frac{1}{\sqrt{x}+x} dx \)
Answer: Solution: Let \( I = \int \frac{1}{\sqrt{x}+x} dx \) \( = \int \frac{1}{\sqrt{x}(1+\sqrt{x})} dx \) \( = \int \frac{1/\sqrt{x}}{1+\sqrt{x}} dx \) (This intermediate step is slightly off from the OCR, which had `∫(1/√x) / (1+√x) dx`. I will keep the OCR structure) \( = \int \frac{1/\sqrt{x}}{1+\sqrt{x}} dx \) Put \( 1 + \sqrt{x} = t \)
\( \implies \frac{1}{2\sqrt{x}} dx = dt \)
\( \implies \frac{1}{\sqrt{x}} dx = 2 dt \)
\( \implies I = \int \frac{1}{t} \cdot (2 dt) = 2\int \frac{1}{t} dt \) \( = 2 \log |t| + c = 2 \log |1 + \sqrt{x}| + c. \)
In simple words: This integral is solved by factoring \( \sqrt{x} \) from the denominator, leading to \( \sqrt{x}(1+\sqrt{x}) \). Then, substituting \( t = 1+\sqrt{x} \) makes its derivative \( \frac{1}{2\sqrt{x}} dx \) align perfectly with the remaining terms for integration.

🎯 Exam Tip: When dealing with expressions involving \( x \) and \( \sqrt{x} \), try factoring out \( \sqrt{x} \). This often simplifies the denominator and creates a form suitable for substitution, typically with \( t = \sqrt{x} \) or an expression containing it.

Question 10. \( \int \frac{1}{x(x^6+1)} dx \)
Answer: Solution: Let \( I = \int \frac{1}{x(x^6+1)} dx \) \( = \int \frac{x^5}{x^6(x^6+1)} dx \) Put \( x^6 = t \)
\( \implies 6x^5dx = dt \)
\( \implies x^5dx = \frac{dt}{6} \)
\( \implies I = \int \frac{1}{t(t+1)} \cdot \frac{dt}{6} \) \( = \frac{1}{6} \int \frac{(t+1)-t}{t(t+1)} dt \) \( = \frac{1}{6} \int \left[ \frac{1}{t} - \frac{1}{t+1} \right] dt \) \( = \frac{1}{6} \left[ \int \frac{1}{t} dt - \int \frac{1}{t+1} dt \right] \) \( = \frac{1}{6} [ \log |t| - \log |t+1| ] + c \) \( = \frac{1}{6} \log \left| \frac{t}{t+1} \right| + c \) \( = \frac{1}{6} \log \left| \frac{x^6}{x^6+1} \right| + c. \)
In simple words: This integral is solved by multiplying the numerator and denominator by \( x^5 \) to prepare for substitution. Setting \( t = x^6 \) simplifies the expression, which is then solved using partial fractions or by algebraic manipulation to integrate logarithmic terms.

🎯 Exam Tip: When faced with an integral like \( \int \frac{1}{x(x^n+1)} dx \), try multiplying the numerator and denominator by \( x^{n-1} \) to make the substitution \( t = x^n \) feasible. This often leads to a partial fraction decomposition or a direct logarithmic form.