Maharashtra Board Class 12 Maths Part 1 Chapter 5 Integration 5.4 Solutions

Evaluate The Following.

Question 1. \( \int \frac{1}{4x^2-1} dx \)
Answer: Solution:
\( \int \frac{1}{4x^2-1} dx = \int \frac{1}{4(x^2-\frac{1}{4})} dx \)
\( = \frac{1}{4} \int \frac{1}{x^2-(\frac{1}{2})^2} dx \)
\( = \frac{1}{4} \times \frac{1}{2(\frac{1}{2})} \log \left| \frac{x-\frac{1}{2}}{x+\frac{1}{2}} \right| + c \)
\( = \frac{1}{4} \log \left| \frac{2x-1}{2x+1} \right| + c. \)
In simple words: This problem involves integrating a rational function. We factor out the constant from the denominator and then use the standard integral formula for \( \int \frac{1}{x^2-a^2} dx \), simplifying the logarithmic term.

🎯 Exam Tip: Remember to factor out coefficients of \( x^2 \) to make the denominator conform to standard integral forms like \( x^2-a^2 \). Pay close attention to algebraic simplification, especially within the logarithm.

Question 2. \( \int \frac{1}{x^2+4x-5} dx \)
Answer: Solution:
\( \int \frac{1}{x^2+4x-5} dx \)
\( = \int \frac{1}{(x^2+4x+4)-4-5} dx \)
\( = \int \frac{1}{(x+2)^2 - (3)^2} dx \)
\( = \frac{1}{2 \times 3} \log \left| \frac{x+2-3}{x+2+3} \right| + c \)
\( = \frac{1}{6} \log \left| \frac{x-1}{x+5} \right| + c. \)
In simple words: To solve this integral, we complete the square in the denominator to transform it into the form \( (u^2 - a^2) \). Then, we apply the standard integration formula for \( \int \frac{1}{u^2-a^2} du \) and simplify.

🎯 Exam Tip: Completing the square is a crucial technique for quadratic denominators. Ensure the constant term added and subtracted correctly maintains the expression's value, and then apply the appropriate formula for \( \int \frac{1}{x^2-a^2} dx \) or \( \int \frac{1}{a^2-x^2} dx \).

Question 3. \( \int \frac{1}{4x^2-20x+17} dx \)
Answer: Solution:
\( \int \frac{1}{4x^2-20x+17} dx \)
\( = \frac{1}{4} \int \frac{1}{x^2-5x+\frac{17}{4}} dx \)
\( = \frac{1}{4} \int \frac{1}{x^2-5x+\frac{25}{4}-\frac{25}{4}+\frac{17}{4}} dx \)
\( = \frac{1}{4} \int \frac{1}{(x-\frac{5}{2})^2 - (\frac{\sqrt{2}}{2})^2} dx \)
\( = \frac{1}{4} \times \frac{1}{2(\frac{\sqrt{2}}{2})} \log \left| \frac{x-\frac{5}{2}-\frac{\sqrt{2}}{2}}{x-\frac{5}{2}+\frac{\sqrt{2}}{2}} \right| + c \)
\( = \frac{1}{4\sqrt{2}} \log \left| \frac{2x-5-\sqrt{2}}{2x-5+\sqrt{2}} \right| + c. \)
In simple words: First, factor out the coefficient of \( x^2 \) from the denominator. Then, complete the square for the quadratic expression and apply the standard integral formula \( \int \frac{1}{u^2-a^2} du \), simplifying the resulting logarithmic expression.

🎯 Exam Tip: When the coefficient of \( x^2 \) is not 1, factor it out before completing the square. Be meticulous with fractions and square roots when applying the integral formula and simplifying the final expression.

Question 4. \( \int \frac{x}{4x^4-20x^2-3} dx \)
Answer: Solution:
Let \( I = \int \frac{x}{4x^4-20x^2-3} dx \)
Put \( x^2 = t \)
\( \therefore 2x dx = dt \)
\( \therefore x dx = \frac{dt}{2} \)
\( \therefore I = \int \frac{1}{4t^2-20t-3} \frac{dt}{2} \)
\( = \frac{1}{2} \times \frac{1}{4} \int \frac{1}{t^2-5t-\frac{3}{4}} dt \)
\( = \frac{1}{8} \int \frac{1}{t^2-5t+\frac{25}{4}-\frac{25}{4}-\frac{3}{4}} dt \)
\( = \frac{1}{8} \int \frac{1}{(t-\frac{5}{2})^2 - (\sqrt{7})^2} dt \)
\( = \frac{1}{8} \times \frac{1}{2\sqrt{7}} \log \left| \frac{t-\frac{5}{2}-\sqrt{7}}{t-\frac{5}{2}+\sqrt{7}} \right| + c \)
\( = \frac{1}{16\sqrt{7}} \log \left| \frac{2t-5-2\sqrt{7}}{2t-5+2\sqrt{7}} \right| + c \)
\( = \frac{1}{16\sqrt{7}} \log \left| \frac{2x^2-5-2\sqrt{7}}{2x^2-5+2\sqrt{7}} \right| + c. \)
In simple words: This integral requires a substitution method first, letting \( t = x^2 \). After substitution, complete the square in the denominator for the resulting quadratic in \( t \). Finally, apply the standard integral formula for \( \int \frac{1}{u^2-a^2} du \) and substitute back for \( x \).

🎯 Exam Tip: For integrals with \( x \) and \( x^4 \) terms, try the substitution \( t = x^2 \). Remember to also differentiate \( x^2 \) to change the \( dx \) term. Don't forget to substitute back the original variable at the end.

Question 5. \( \int \frac{x^3}{16x^8-25} dx \)
Answer: Solution:
Let \( I = \int \frac{x^3}{16x^8-25} dx \)
Put \( x^4 = t \)
\( \therefore 4x^3 dx = dt \)
\( \therefore x^3 dx = \frac{dt}{4} \)
\( \therefore I = \int \frac{1}{16t^2-25} \frac{dt}{4} \)
\( = \frac{1}{4} \times \frac{1}{16} \int \frac{1}{t^2-\frac{25}{16}} dt \)
\( = \frac{1}{64} \int \frac{1}{t^2-(\frac{5}{4})^2} dt \)
\( = \frac{1}{64} \times \frac{1}{2 \times \frac{5}{4}} \log \left| \frac{t-\frac{5}{4}}{t+\frac{5}{4}} \right| + c \)
\( = \frac{1}{64} \times \frac{4}{10} \log \left| \frac{4t-5}{4t+5} \right| + c \)
\( = \frac{1}{160} \log \left| \frac{4x^4-5}{4x^4+5} \right| + c. \)
In simple words: Use the substitution method by letting \( t = x^4 \), which simplifies the integrand. Then, factor out the constant from the denominator, apply the standard integral formula for \( \int \frac{1}{u^2-a^2} du \), and replace \( t \) with \( x^4 \) in the final answer.

🎯 Exam Tip: For integrals with terms like \( x^3 \) and \( x^8 \), consider a substitution like \( t = x^4 \) to simplify the integrand. Careful algebraic manipulation of constants and fractions is key to arriving at the correct logarithmic form.

Question 6. \( \int \frac{1}{a^2-b^2x^2} dx \)
Answer: Solution:
\( \int \frac{1}{a^2-b^2x^2} dx \)
\( = \int \frac{1}{b^2(\frac{a^2}{b^2}-x^2)} dx \)
\( = \frac{1}{b^2} \int \frac{1}{(\frac{a}{b})^2-x^2} dx \)
\( = \frac{1}{b^2} \times \frac{1}{2(\frac{a}{b})} \log \left| \frac{\frac{a}{b}+x}{\frac{a}{b}-x} \right| + c \)
\( = \frac{1}{2ab} \log \left| \frac{a+bx}{a-bx} \right| + c. \)
In simple words: Begin by factoring out \( b^2 \) from the denominator to make the \( x^2 \) term have a coefficient of 1. This converts the expression into the form \( \int \frac{1}{A^2-x^2} dx \), allowing for direct application of the standard integral formula.

🎯 Exam Tip: When integrating forms with \( a^2-b^2x^2 \), always factor out \( b^2 \) to isolate \( x^2 \) and identify \( A = \frac{a}{b} \). Ensure careful algebraic simplification to reach the compact final logarithmic form.

Question 7. \( \int \frac{1}{7+6x-x^2} dx \)
Answer: Solution:
\( \int \frac{1}{7+6x-x^2} dx \)
\( = \int \frac{1}{7-(x^2-6x-9)+9} dx \) (Correcting the sign in (x^2-6x+9))
\( = \int \frac{1}{7-(-(-9)+9-(x^2-6x))} dx \) (Re-evaluating the completion of square)
Let's re-do the completion of square carefully for \( -x^2+6x+7 \):
\( -(x^2-6x-7) \)
\( -(x^2-6x+9-9-7) \)
\( -((x-3)^2-16) \)
\( 16-(x-3)^2 \)
So, \( \int \frac{1}{7+6x-x^2} dx = \int \frac{1}{16-(x-3)^2} dx \)
\( = \int \frac{1}{(4)^2 - (x-3)^2} dx \)
\( = \frac{1}{2 \times 4} \log \left| \frac{4+(x-3)}{4-(x-3)} \right| + c \)
\( = \frac{1}{8} \log \left| \frac{4+x-3}{4-x+3} \right| + c \)
\( = \frac{1}{8} \log \left| \frac{x+1}{7-x} \right| + c. \)
In simple words: For a quadratic denominator with a negative \( x^2 \) term, first factor out the negative sign and then complete the square for the quadratic. This will transform the integral into the form \( \int \frac{1}{a^2-u^2} du \), allowing for direct application of the corresponding integral formula.

🎯 Exam Tip: When the \( x^2 \) term is negative, factor out the negative sign before completing the square to get a form like \( a^2-u^2 \). This avoids errors in applying the formula \( \int \frac{1}{a^2-u^2} du \), which is different from \( \int \frac{1}{u^2-a^2} du \).

Question 8. \( \int \frac{1}{\sqrt{3x^2+8}} dx \)
Answer: Solution:
\( \int \frac{1}{\sqrt{3x^2+8}} dx \)
\( = \int \frac{1}{\sqrt{3(x^2+\frac{8}{3})}} dx \)
\( = \frac{1}{\sqrt{3}} \int \frac{1}{\sqrt{x^2+(\frac{\sqrt{8}}{\sqrt{3}})^2}} dx \)
\( = \frac{1}{\sqrt{3}} \log \left| x + \sqrt{x^2+(\frac{\sqrt{8}}{\sqrt{3}})^2} \right| + c \)
\( = \frac{1}{\sqrt{3}} \log \left| x + \sqrt{x^2+\frac{8}{3}} \right| + c \)
\( = \frac{1}{\sqrt{3}} \log \left| x + \frac{\sqrt{3x^2+8}}{\sqrt{3}} \right| + c \) (Alternative simplification shown in OCR output using \( \sqrt{3x} \) form, but direct formula application is typically with \( x \))
Let's stick to the standard formula:
\( \int \frac{1}{\sqrt{ax^2+b}} dx = \frac{1}{\sqrt{a}} \log \left| x + \sqrt{x^2+\frac{b}{a}} \right| + C \)
Applying this:
\( = \frac{1}{\sqrt{3}} \log \left| x + \sqrt{x^2+\frac{8}{3}} \right| + c \)
\( = \frac{1}{\sqrt{3}} \log \left| \frac{ \sqrt{3}x + \sqrt{3x^2+8} }{\sqrt{3}} \right| + c \)
\( = \frac{1}{\sqrt{3}} \left( \log \left| \sqrt{3}x + \sqrt{3x^2+8} \right| - \log(\sqrt{3}) \right) + c \)
Since \( -\frac{1}{\sqrt{3}}\log(\sqrt{3}) \) is a constant, it can be absorbed into \( c \).
So, \( = \frac{1}{\sqrt{3}} \log \left| \sqrt{3}x + \sqrt{3x^2+8} \right| + c. \)
OCR's solution directly uses \( \log|\sqrt{3}x + \sqrt{(\sqrt{3}x)^2+(\sqrt{8})^2}| \):
\( \int \frac{1}{\sqrt{3x^2+8}} dx = \int \frac{1}{\sqrt{(\sqrt{3}x)^2+(\sqrt{8})^2}} dx \)
Let \( u = \sqrt{3}x \), then \( du = \sqrt{3} dx \implies dx = \frac{du}{\sqrt{3}} \).
\( = \int \frac{1}{\sqrt{u^2+(\sqrt{8})^2}} \frac{du}{\sqrt{3}} \)
\( = \frac{1}{\sqrt{3}} \log \left| u + \sqrt{u^2+(\sqrt{8})^2} \right| + c \)
\( = \frac{1}{\sqrt{3}} \log \left| \sqrt{3}x + \sqrt{(\sqrt{3}x)^2+(\sqrt{8})^2} \right| + c \)
\( = \frac{1}{\sqrt{3}} \log \left| \sqrt{3}x + \sqrt{3x^2+8} \right| + c. \)
In simple words: This integral is of the form \( \int \frac{1}{\sqrt{ax^2+b}} dx \). We use the substitution \( u = \sqrt{a}x \) to convert it into a standard form \( \int \frac{1}{\sqrt{u^2+A^2}} du \), and then apply the logarithmic integral formula.

🎯 Exam Tip: For integrals with \( \sqrt{ax^2+b} \), consider a substitution like \( u = \sqrt{a}x \) to simplify the radical. Remember to adjust the \( dx \) term accordingly and apply the formula \( \int \frac{1}{\sqrt{x^2+a^2}} dx = \log|x+\sqrt{x^2+a^2}| + C \).

Question 9. \( \int \frac{1}{\sqrt{x^2+4x+29}} dx \)
Answer: Solution:
\( \int \frac{1}{\sqrt{x^2+4x+29}} dx \)
\( = \int \frac{1}{\sqrt{(x^2+4x+4)-4+29}} dx \)
\( = \int \frac{1}{\sqrt{(x+2)^2+25}} dx \)
\( = \int \frac{1}{\sqrt{(x+2)^2+(5)^2}} dx \)
\( = \log \left| (x+2) + \sqrt{(x+2)^2+(5)^2} \right| + c \)
\( = \log \left| (x+2) + \sqrt{x^2+4x+29} \right| + c. \)
In simple words: Complete the square for the quadratic expression under the square root in the denominator. This transforms the integral into the standard form \( \int \frac{1}{\sqrt{u^2+a^2}} du \), which integrates to \( \log|u+\sqrt{u^2+a^2}| \).

🎯 Exam Tip: Completing the square is essential for quadratic expressions inside square roots. After completing the square, the integral will match one of the standard forms, usually \( \int \frac{1}{\sqrt{x^2 \pm a^2}} dx \).

Question 10. \( \int \frac{1}{\sqrt{3x^2-5}} dx \)
Answer: Solution:
\( \int \frac{1}{\sqrt{3x^2-5}} dx \)
\( = \int \frac{1}{\sqrt{3(x^2-\frac{5}{3})}} dx \)
\( = \frac{1}{\sqrt{3}} \int \frac{1}{\sqrt{x^2-(\frac{\sqrt{5}}{\sqrt{3}})^2}} dx \)
\( = \frac{1}{\sqrt{3}} \log \left| x + \sqrt{x^2-(\frac{\sqrt{5}}{\sqrt{3}})^2} \right| + c \)
\( = \frac{1}{\sqrt{3}} \log \left| x + \sqrt{x^2-\frac{5}{3}} \right| + c \)
\( = \frac{1}{\sqrt{3}} \log \left| \frac{\sqrt{3}x + \sqrt{3x^2-5}}{\sqrt{3}} \right| + c \)
\( = \frac{1}{\sqrt{3}} \log \left| \sqrt{3}x + \sqrt{3x^2-5} \right| + c. \)
In simple words: First, factor out the coefficient of \( x^2 \) from under the square root. Then, the integral takes the form \( \int \frac{1}{\sqrt{u^2-a^2}} du \), which can be directly integrated using the formula \( \log|u+\sqrt{u^2-a^2}| \), remembering to substitute back.

🎯 Exam Tip: When dealing with \( \sqrt{ax^2 \pm b} \), factor out \( \sqrt{a} \) to simplify to a standard \( \sqrt{x^2 \pm A^2} \) form. Be careful with fractional constants in the \( a^2 \) term when applying the logarithmic formula.

Question 11. \( \int \frac{1}{\sqrt{x^2-8x-20}} dx \)
Answer: Solution:
\( \int \frac{1}{\sqrt{x^2-8x-20}} dx \)
\( = \int \frac{1}{\sqrt{(x^2-8x+16)-16-20}} dx \)
\( = \int \frac{1}{\sqrt{(x-4)^2-36}} dx \)
\( = \int \frac{1}{\sqrt{(x-4)^2-(6)^2}} dx \)
\( = \log \left| (x-4) + \sqrt{(x-4)^2-(6)^2} \right| + c \)
\( = \log \left| (x-4) + \sqrt{x^2-8x-20} \right| + c. \)
In simple words: Complete the square for the quadratic expression under the square root. This transforms the integral into the standard form \( \int \frac{1}{\sqrt{u^2-a^2}} du \), which integrates to \( \log|u+\sqrt{u^2-a^2}| \).

🎯 Exam Tip: Always start by completing the square for quadratic expressions under the radical. This converts the integral into one of the three standard forms involving square roots, allowing for straightforward application of the respective formula.