Samacheer Kalvi Class 12 Business Maths Solutions Chapter 2 Integral Calculus I More Ques

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 2 Integral Calculus I Miscellaneous Problems

 

Question 1. \( \int\frac { 1 }{\sqrt{x+2}-\sqrt{x+3}} dx \)
Answer: We begin by simplifying the expression in the integral. First, we need to conjugate the denominator to remove the square roots from the bottom part.
\( \int \frac { 1 }{ \sqrt{x+2}-\sqrt{x+3} } \times \frac { \sqrt{x+2}+\sqrt{x+3} }{ \sqrt{x+2}+\sqrt{x+3} } dx \)
\( = \int \frac { \sqrt{x+2}+\sqrt{x+3} }{ (x+2)-(x+3) } dx \)
\( = \int \frac { \sqrt{x+2}+\sqrt{x+3} }{ -1 } dx \)
\( = - \int ( \sqrt{x+2}+\sqrt{x+3} ) dx \)
\( = - \left[ \int (x+2)^{1/2} dx + \int (x+3)^{1/2} dx \right] \)
\( = - \left[ \frac { (x+2)^{3/2} }{ 3/2 } + \frac { (x+3)^{3/2} }{ 3/2 } \right] + C \)
\( = - \frac { 2 }{ 3 } \left[ (x+2)^{3/2} + (x+3)^{3/2} \right] + C \)
The final integrated expression is obtained by applying the power rule for integration to each term.
In simple words: To solve this, we multiplied the top and bottom by the conjugate of the denominator, which helped us get rid of the square roots at the bottom. Then, we integrated each part separately using the power rule.

🎯 Exam Tip: Always remember to conjugate the denominator when you have a sum or difference of square roots in the denominator of an integrand. This step simplifies the expression for easier integration.

 

Question 2. \( \int\frac { dx }{2-3x-2x^2} \)
Answer: First, we rewrite the denominator in a standard quadratic form and factor it.
\( \int \frac { dx }{ -2x^2-3x+2 } = \int \frac { dx }{ -(2x^2+3x-2) } \)
\( = - \int \frac { dx }{ (2x-1)(x+2) } \)
Now, we use partial fraction decomposition to break the fraction into simpler terms.
Let \( \frac { 1 }{ (2x-1)(x+2) } = \frac { A }{ 2x-1 } + \frac { B }{ x+2 } \)
\( 1 = A(x+2) + B(2x-1) \)
To find A, set \( 2x-1=0 \implies x = \frac{1}{2} \).
\( 1 = A\left(\frac{1}{2}+2\right) + B(0) \)
\( 1 = A\left(\frac{5}{2}\right) \implies A = \frac{2}{5} \)
To find B, set \( x+2=0 \implies x = -2 \).
\( 1 = A(0) + B(2(-2)-1) \)
\( 1 = B(-5) \implies B = -\frac{1}{5} \)
Substitute A and B back into the integral:
\( = - \int \left( \frac { 2/5 }{ 2x-1 } - \frac { 1/5 }{ x+2 } \right) dx \)
\( = - \frac{2}{5} \int \frac{1}{2x-1} dx + \frac{1}{5} \int \frac{1}{x+2} dx \)
\( = - \frac{2}{5} \cdot \frac{1}{2} \log|2x-1| + \frac{1}{5} \log|x+2| + C \)
\( = - \frac{1}{5} \log|2x-1| + \frac{1}{5} \log|x+2| + C \)
\( = \frac{1}{5} (\log|x+2| - \log|2x-1|) + C \)
\( = \frac{1}{5} \log \left| \frac{x+2}{2x-1} \right| + C \)
This method of partial fractions makes integrating rational functions much simpler.
In simple words: First, we arranged the bottom part of the fraction. Then, we split the complex fraction into two simpler ones using a method called partial fractions. After that, we integrated each simple fraction to get the final answer.

🎯 Exam Tip: When dealing with integrals of rational functions, always check if the denominator can be factored. If it can, partial fraction decomposition is often the most effective technique. Remember to handle the negative sign carefully during factorization and integration.

 

Question 3. \( \int\frac { dx }{e^x+6+5e^{-x}} \)
Answer: To solve this integral, we use a substitution to simplify the expression.
First, rewrite the denominator:
\( \int \frac { dx }{ e^x+6+5e^{-x} } = \int \frac { dx }{ e^x+6+\frac{5}{e^x} } \)
Multiply numerator and denominator by \( e^x \):
\( = \int \frac { e^x dx }{ (e^x)^2+6e^x+5 } \)
Let \( t = e^x \).
Then \( dt = e^x dx \).
Substitute these into the integral:
\( = \int \frac { dt }{ t^2+6t+5 } \)
Factor the denominator:
\( = \int \frac { dt }{ (t+1)(t+5) } \)
Now, use partial fraction decomposition:
Let \( \frac { 1 }{ (t+1)(t+5) } = \frac { A }{ t+1 } + \frac { B }{ t+5 } \)
\( 1 = A(t+5) + B(t+1) \)
To find A, set \( t+1=0 \implies t = -1 \).
\( 1 = A(-1+5) + B(0) \)
\( 1 = 4A \implies A = \frac{1}{4} \)
To find B, set \( t+5=0 \implies t = -5 \).
\( 1 = A(0) + B(-5+1) \)
\( 1 = -4B \implies B = -\frac{1}{4} \)
Substitute A and B back into the integral:
\( = \int \left( \frac { 1/4 }{ t+1 } - \frac { 1/4 }{ t+5 } \right) dt \)
\( = \frac{1}{4} \int \frac{1}{t+1} dt - \frac{1}{4} \int \frac{1}{t+5} dt \)
\( = \frac{1}{4} \log|t+1| - \frac{1}{4} \log|t+5| + C \)
\( = \frac{1}{4} \log \left| \frac{t+1}{t+5} \right| + C \)
Finally, substitute back \( t = e^x \):
\( = \frac{1}{4} \log \left| \frac{e^x+1}{e^x+5} \right| + C \)
Substitution helps convert complex exponential integrals into simpler algebraic forms, making them easier to solve.
In simple words: We first changed \( e^{-x} \) to \( \frac{1}{e^x} \) and then replaced \( e^x \) with 't'. This made the integral look like a simple fraction, which we then split into two parts. After integrating these parts, we put \( e^x \) back in place of 't'.

🎯 Exam Tip: For integrals involving \( e^x \) and \( e^{-x} \), a common and effective strategy is to multiply the numerator and denominator by \( e^x \) (or \( e^{-x} \)) and then use the substitution \( t = e^x \). This transforms the integral into a rational function, which can often be solved using partial fractions.

 

Question 4. \( \int\sqrt {2x^2-3} dx \)
Answer: We need to integrate the expression \( \sqrt{2x^2-3} \).
First, rewrite the expression to fit the standard integral form \( \sqrt{x^2-a^2} \).
\( \int \sqrt{2x^2-3} dx = \int \sqrt{(\sqrt{2}x)^2 - (\sqrt{3})^2} dx \)
We know the formula for \( \int \sqrt{X^2-A^2} dX = \frac{X}{2}\sqrt{X^2-A^2} - \frac{A^2}{2}\log|X+\sqrt{X^2-A^2}| + C \).
Here, \( X = \sqrt{2}x \) and \( A = \sqrt{3} \).
Also, for the term \( dX \), we have \( d(\sqrt{2}x) = \sqrt{2} dx \), so \( dx = \frac{1}{\sqrt{2}} dX \).
Substitute these into the formula:
\( = \frac{1}{\sqrt{2}} \left[ \frac{\sqrt{2}x}{2}\sqrt{(\sqrt{2}x)^2-(\sqrt{3})^2} - \frac{(\sqrt{3})^2}{2}\log|\sqrt{2}x+\sqrt{(\sqrt{2}x)^2-(\sqrt{3})^2}| \right] + C \)
\( = \frac{1}{\sqrt{2}} \left[ \frac{\sqrt{2}x}{2}\sqrt{2x^2-3} - \frac{3}{2}\log|\sqrt{2}x+\sqrt{2x^2-3}| \right] + C \)
Distribute \( \frac{1}{\sqrt{2}} \) inside:
\( = \frac{x}{2}\sqrt{2x^2-3} - \frac{3}{2\sqrt{2}}\log|\sqrt{2}x+\sqrt{2x^2-3}| + C \)
The final answer simplifies to this form. This integral form is a common application of specific integration formulas.
In simple words: We changed the given expression to match a known integration formula that deals with square roots of terms like \( X^2-A^2 \). Then, we plugged in our values into this formula and simplified the result to get the final answer.

🎯 Exam Tip: When integrating expressions like \( \sqrt{ax^2+b} \), try to transform them into one of the standard integral forms like \( \sqrt{x^2 \pm a^2} \) or \( \sqrt{a^2-x^2} \). Remember to adjust the `dx` term correctly if there's a coefficient with \( x \).

 

Question 5. \( \int\sqrt { 9x^2+12x+3} dx \)
Answer: To solve this integral, we will complete the square inside the square root.
\( I = \int \sqrt{9x^2+12x+3} dx \)
Factor out 9 from the terms under the square root:
\( = \int \sqrt{9(x^2+\frac{12}{9}x+\frac{3}{9})} dx \)
\( = \int \sqrt{9(x^2+\frac{4}{3}x+\frac{1}{3})} dx \)
Take \( \sqrt{9} = 3 \) outside the integral:
\( = 3 \int \sqrt{x^2+\frac{4}{3}x+\frac{1}{3}} dx \)
Complete the square for \( x^2+\frac{4}{3}x \). Add and subtract \( \left(\frac{1}{2} \cdot \frac{4}{3}\right)^2 = \left(\frac{2}{3}\right)^2 = \frac{4}{9} \).
\( = 3 \int \sqrt{x^2+\frac{4}{3}x+\frac{4}{9} - \frac{4}{9} + \frac{1}{3}} dx \)
\( = 3 \int \sqrt{\left(x+\frac{2}{3}\right)^2 - \frac{4}{9} + \frac{3}{9}} dx \)
\( = 3 \int \sqrt{\left(x+\frac{2}{3}\right)^2 - \frac{1}{9}} dx \)
\( = 3 \int \sqrt{\left(x+\frac{2}{3}\right)^2 - \left(\frac{1}{3}\right)^2} dx \)
This is in the form \( \int \sqrt{X^2-A^2} dX \), where \( X = x+\frac{2}{3} \) and \( A = \frac{1}{3} \).
The formula is \( \frac{X}{2}\sqrt{X^2-A^2} - \frac{A^2}{2}\log|X+\sqrt{X^2-A^2}| + C \).
Substitute \( X \) and \( A \) into the formula:
\( = 3 \left[ \frac{x+\frac{2}{3}}{2}\sqrt{\left(x+\frac{2}{3}\right)^2-\left(\frac{1}{3}\right)^2} - \frac{\left(\frac{1}{3}\right)^2}{2}\log\left|x+\frac{2}{3}+\sqrt{\left(x+\frac{2}{3}\right)^2-\left(\frac{1}{3}\right)^2}\right| \right] + C \)
Simplify the terms:
\( = 3 \left[ \frac{3x+2}{6}\sqrt{x^2+\frac{4}{3}x+\frac{1}{3}} - \frac{1/9}{2}\log\left|x+\frac{2}{3}+\sqrt{x^2+\frac{4}{3}x+\frac{1}{3}}\right| \right] + C \)
\( = 3 \left[ \frac{3x+2}{6} \cdot \frac{1}{3}\sqrt{9x^2+12x+3} - \frac{1}{18}\log\left|x+\frac{2}{3}+\frac{1}{3}\sqrt{9x^2+12x+3}\right| \right] + C \)
\( = \frac{3x+2}{6}\sqrt{9x^2+12x+3} - \frac{1}{6}\log\left|x+\frac{2}{3}+\frac{1}{3}\sqrt{9x^2+12x+3}\right| + C \)
This method of completing the square is essential for many integration problems involving quadratic expressions under a square root.
In simple words: We began by changing the expression inside the square root to make it simpler, using a trick called "completing the square." This helped us match it to a known integration rule for square roots. Then, we applied the rule and simplified the terms to get our final answer.

🎯 Exam Tip: For integrals of the form \( \int \sqrt{ax^2+bx+c} dx \), the crucial first step is to complete the square inside the square root. This converts the quadratic expression into the form \( X^2 \pm A^2 \) or \( A^2 - X^2 \), which can then be integrated using standard formulas. Be careful with factoring out coefficients and algebraic simplifications.

 

Question 6. \( \int(x + 1)^2 \log x dx \)
Answer: We use integration by parts, which is given by \( \int u dv = uv - \int v du \).
We choose \( u = \log x \) and \( dv = (x+1)^2 dx \).
Then, we find \( du \) and \( v \).
\( du = \frac{1}{x} dx \)
\( v = \int (x+1)^2 dx = \frac{(x+1)^3}{3} \)
Now, apply the integration by parts formula:
\( \int (x+1)^2 \log x dx = (\log x) \frac{(x+1)^3}{3} - \int \frac{(x+1)^3}{3} \cdot \frac{1}{x} dx \)
\( = \frac{(x+1)^3}{3} \log x - \frac{1}{3} \int \frac{x^3+3x^2+3x+1}{x} dx \)
\( = \frac{(x+1)^3}{3} \log x - \frac{1}{3} \int \left( x^2+3x+3+\frac{1}{x} \right) dx \)
Integrate each term in the second part:
\( = \frac{(x+1)^3}{3} \log x - \frac{1}{3} \left[ \frac{x^3}{3}+\frac{3x^2}{2}+3x+\log|x| \right] + C \)
\( = \frac{(x+1)^3}{3} \log x - \frac{x^3}{9}-\frac{x^2}{2}-x-\frac{1}{3}\log|x| + C \)
The key to successful integration by parts is choosing `u` and `dv` correctly using the LIATE rule (Logarithmic, Inverse trig, Algebraic, Trigonometric, Exponential).
In simple words: We solved this integral using a method called "integration by parts." We split the problem into two parts, \( \log x \) and \( (x+1)^2 \), and then used a special formula to integrate them. After expanding and integrating the remaining terms, we got the final answer.

🎯 Exam Tip: When using integration by parts, the choice of \( u \) and \( dv \) is critical. A good heuristic is the LIATE rule: choose \( u \) as the function that comes first in the order of Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential. For \( \log x \) and \( (x+1)^2 \), \( \log x \) should be \( u \).

 

Question 7. \( \int\log (x-\sqrt {x^2-1}) dx \)
Answer: We will use integration by parts, \( \int u dv = uv - \int v du \).
Let \( u = \log(x-\sqrt{x^2-1}) \) and \( dv = dx \).
Then \( v = \int dx = x \).
Now, we need to find \( du \).
\( \frac{du}{dx} = \frac{1}{x-\sqrt{x^2-1}} \cdot \frac{d}{dx}(x-\sqrt{x^2-1}) \)
\( = \frac{1}{x-\sqrt{x^2-1}} \cdot \left( 1 - \frac{1}{2\sqrt{x^2-1}} \cdot 2x \right) \)
\( = \frac{1}{x-\sqrt{x^2-1}} \cdot \left( 1 - \frac{x}{\sqrt{x^2-1}} \right) \)
\( = \frac{1}{x-\sqrt{x^2-1}} \cdot \left( \frac{\sqrt{x^2-1}-x}{\sqrt{x^2-1}} \right) \)
\( = \frac{1}{x-\sqrt{x^2-1}} \cdot \frac{-(x-\sqrt{x^2-1})}{\sqrt{x^2-1}} \)
\( = - \frac{1}{\sqrt{x^2-1}} \)
So, \( du = - \frac{1}{\sqrt{x^2-1}} dx \).
Apply the integration by parts formula:
\( \int \log(x-\sqrt{x^2-1}) dx = x \log(x-\sqrt{x^2-1}) - \int x \left( - \frac{1}{\sqrt{x^2-1}} \right) dx \)
\( = x \log(x-\sqrt{x^2-1}) + \int \frac{x}{\sqrt{x^2-1}} dx \)
For the second integral, let \( t = x^2-1 \), so \( dt = 2x dx \implies x dx = \frac{1}{2} dt \).
\( \int \frac{x}{\sqrt{x^2-1}} dx = \int \frac{1}{\sqrt{t}} \frac{1}{2} dt = \frac{1}{2} \int t^{-1/2} dt \)
\( = \frac{1}{2} \frac{t^{1/2}}{1/2} + C = t^{1/2} + C = \sqrt{x^2-1} + C \)
Substitute this back into the main equation:
\( = x \log(x-\sqrt{x^2-1}) + \sqrt{x^2-1} + C \)
This integral shows how careful differentiation of a logarithmic function can lead to a simplified integrand.
In simple words: We used integration by parts. We picked the logarithm part as 'u' and 'dx' as 'dv'. Then, we found 'du' by differentiating the logarithm carefully. After plugging these parts into the integration formula, the problem became much simpler, and we could integrate the remaining part directly.

🎯 Exam Tip: When integrating \( \log(f(x)) \), choosing \( u=\log(f(x)) \) and \( dv=dx \) is often effective. Pay close attention to the differentiation of \( \log(x-\sqrt{x^2-1}) \), as it simplifies significantly, making the \( \int v du \) term manageable.

 

Question 8. \( \int_{0}^{1} \sqrt { x(x-1)} dx \)
Answer: We need to evaluate the definite integral \( \int_{0}^{1} \sqrt{x(x-1)} dx \).
First, simplify the expression inside the square root:
\( \sqrt{x(x-1)} = \sqrt{x^2-x} \)
Next, complete the square for \( x^2-x \):
\( x^2-x = x^2-x+\left(\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^2 = \left(x-\frac{1}{2}\right)^2 - \left(\frac{1}{2}\right)^2 \)
So the integral becomes:
\( \int_{0}^{1} \sqrt{\left(x-\frac{1}{2}\right)^2 - \left(\frac{1}{2}\right)^2} dx \)
This is in the form \( \int \sqrt{X^2-A^2} dX \), where \( X = x-\frac{1}{2} \) and \( A = \frac{1}{2} \).
The formula for this indefinite integral is \( \frac{X}{2}\sqrt{X^2-A^2} - \frac{A^2}{2}\log|X+\sqrt{X^2-A^2}| \).
Let's apply the limits of integration.
\( \left[ \frac{x-\frac{1}{2}}{2}\sqrt{\left(x-\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^2} - \frac{\left(\frac{1}{2}\right)^2}{2}\log\left|x-\frac{1}{2}+\sqrt{\left(x-\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^2}\right| \right]_{0}^{1} \)
Substitute \( x-\frac{1}{2} \) with \( X \) and \( \left(x-\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^2 \) with \( x^2-x \):
\( \left[ \frac{x-\frac{1}{2}}{2}\sqrt{x^2-x} - \frac{1/4}{2}\log\left|x-\frac{1}{2}+\sqrt{x^2-x}\right| \right]_{0}^{1} \)
\( \left[ \frac{x-\frac{1}{2}}{2}\sqrt{x^2-x} - \frac{1}{8}\log\left|x-\frac{1}{2}+\sqrt{x^2-x}\right| \right]_{0}^{1} \)
Now, substitute the upper limit \( x=1 \):
\( \text{At } x=1: \frac{1-\frac{1}{2}}{2}\sqrt{1^2-1} - \frac{1}{8}\log\left|1-\frac{1}{2}+\sqrt{1^2-1}\right| \)
\( = \frac{1/2}{2}\sqrt{0} - \frac{1}{8}\log\left|\frac{1}{2}+\sqrt{0}\right| \)
\( = 0 - \frac{1}{8}\log\left|\frac{1}{2}\right| = - \frac{1}{8} \log\left(\frac{1}{2}\right) \)
Substitute the lower limit \( x=0 \):
\( \text{At } x=0: \frac{0-\frac{1}{2}}{2}\sqrt{0^2-0} - \frac{1}{8}\log\left|0-\frac{1}{2}+\sqrt{0^2-0}\right| \)
\( = \frac{-1/2}{2}\sqrt{0} - \frac{1}{8}\log\left|-\frac{1}{2}+\sqrt{0}\right| \)
\( = 0 - \frac{1}{8}\log\left|-\frac{1}{2}\right| = - \frac{1}{8} \log\left(\frac{1}{2}\right) \)
Subtract the lower limit value from the upper limit value:
\( = \left( - \frac{1}{8} \log\left(\frac{1}{2}\right) \right) - \left( - \frac{1}{8} \log\left(\frac{1}{2}\right) \right) = 0 \)
The integral evaluates to zero. This is an important property of some definite integrals.
In simple words: We first rewrote the expression inside the square root by completing the square to match a standard integration formula. Then, we applied the limits of integration (from 0 to 1) to the integrated form. After calculating the values at both limits and subtracting them, the final answer turned out to be zero.

🎯 Exam Tip: When evaluating definite integrals, always simplify the integrand first, often by completing the square if it's a quadratic under a square root. Be careful when substituting the limits, especially with logarithmic terms, and ensure all calculations for both upper and lower limits are precise.

 

Question 9. \( \int_{-1}^{1} x^2 e^{-2x} dx \)
Answer: We need to evaluate the definite integral \( \int_{-1}^{1} x^2 e^{-2x} dx \).
We use generalized integration by parts formula: \( \int u dv = uv - u'v_1 + u''v_2 - u'''v_3 + \dots \)
Here, we choose \( u = x^2 \) (algebraic term) and \( dv = e^{-2x} dx \) (exponential term).

Apply the formula:
\( \int x^2 e^{-2x} dx = x^2 \left( \frac{e^{-2x}}{-2} \right) - (2x) \left( \frac{e^{-2x}}{4} \right) + (2) \left( \frac{e^{-2x}}{-8} \right) - 0 + C \)
\( = - \frac{x^2 e^{-2x}}{2} - \frac{2x e^{-2x}}{4} - \frac{2 e^{-2x}}{8} + C \)
\( = - \frac{x^2 e^{-2x}}{2} - \frac{x e^{-2x}}{2} - \frac{e^{-2x}}{4} + C \)
Now, we evaluate this definite integral from -1 to 1:
\( \left[ - \frac{e^{-2x}}{4} (2x^2+2x+1) \right]_{-1}^{1} \)
Substitute the upper limit \( x=1 \):
\( = - \frac{e^{-2(1)}}{4} (2(1)^2+2(1)+1) = - \frac{e^{-2}}{4} (2+2+1) = - \frac{5e^{-2}}{4} \)
Substitute the lower limit \( x=-1 \):
\( = - \frac{e^{-2(-1)}}{4} (2(-1)^2+2(-1)+1) = - \frac{e^{2}}{4} (2-2+1) = - \frac{e^{2}}{4} \)
Subtract the lower limit value from the upper limit value:
\( = \left( - \frac{5e^{-2}}{4} \right) - \left( - \frac{e^{2}}{4} \right) \)
\( = \frac{e^2 - 5e^{-2}}{4} \)
This problem demonstrates the efficiency of the generalized integration by parts for polynomials multiplied by exponentials.
In simple words: We used a special shortcut for integration by parts, which is helpful when one part is a polynomial and the other is an exponential. We kept taking derivatives of the polynomial and integrals of the exponential. Then, we applied the limits of the integral (from -1 to 1) to find the final numerical answer.

🎯 Exam Tip: For integrals of the form \( \int P(x) e^{ax} dx \) (where \( P(x) \) is a polynomial), the tabular method (generalized integration by parts) is very efficient. Remember to alternate signs (+ - + -) and ensure the derivatives of \( u \) and integrals of \( dv \) are correctly calculated.

 

Question 10. \( \int_{0}^{3} \frac { xdx }{\sqrt {x+1} + \sqrt{5x+1}} \)
Answer: To solve this integral, we first rationalize the denominator by multiplying by its conjugate.
\( I = \int_{0}^{3} \frac{x}{\sqrt{x+1}+\sqrt{5x+1}} dx \)
Multiply numerator and denominator by \( (\sqrt{x+1}-\sqrt{5x+1}) \):
\( = \int_{0}^{3} \frac{x(\sqrt{x+1}-\sqrt{5x+1})}{(\sqrt{x+1}+\sqrt{5x+1})(\sqrt{x+1}-\sqrt{5x+1})} dx \)
\( = \int_{0}^{3} \frac{x(\sqrt{x+1}-\sqrt{5x+1})}{(x+1)-(5x+1)} dx \)
\( = \int_{0}^{3} \frac{x(\sqrt{x+1}-\sqrt{5x+1})}{x+1-5x-1} dx \)
\( = \int_{0}^{3} \frac{x(\sqrt{x+1}-\sqrt{5x+1})}{-4x} dx \)
Cancel \( x \) from numerator and denominator (assuming \( x \neq 0 \), which is true within the integral range except for the lower limit point itself, and the integral still converges):
\( = - \frac{1}{4} \int_{0}^{3} (\sqrt{x+1}-\sqrt{5x+1}) dx \)
\( = - \frac{1}{4} \left[ \int_{0}^{3} (x+1)^{1/2} dx - \int_{0}^{3} (5x+1)^{1/2} dx \right] \)
Integrate each term using the power rule \( \int (ax+b)^n dx = \frac{(ax+b)^{n+1}}{a(n+1)} \):
\( = - \frac{1}{4} \left[ \frac{(x+1)^{3/2}}{3/2} - \frac{(5x+1)^{3/2}}{5 \cdot (3/2)} \right]_{0}^{3} \)
\( = - \frac{1}{4} \left[ \frac{2}{3}(x+1)^{3/2} - \frac{2}{15}(5x+1)^{3/2} \right]_{0}^{3} \)
Now, substitute the limits of integration.
At \( x=3 \):
\( \frac{2}{3}(3+1)^{3/2} - \frac{2}{15}(5(3)+1)^{3/2} = \frac{2}{3}(4)^{3/2} - \frac{2}{15}(16)^{3/2} \)
\( = \frac{2}{3}(8) - \frac{2}{15}(64) = \frac{16}{3} - \frac{128}{15} = \frac{80-128}{15} = -\frac{48}{15} = -\frac{16}{5} \)
At \( x=0 \):
\( \frac{2}{3}(0+1)^{3/2} - \frac{2}{15}(5(0)+1)^{3/2} = \frac{2}{3}(1)^{3/2} - \frac{2}{15}(1)^{3/2} \)
\( = \frac{2}{3} - \frac{2}{15} = \frac{10-2}{15} = \frac{8}{15} \)
Subtract the lower limit value from the upper limit value:
\( = - \frac{1}{4} \left[ \left(-\frac{16}{5}\right) - \left(\frac{8}{15}\right) \right] \)
\( = - \frac{1}{4} \left[ -\frac{48}{15} - \frac{8}{15} \right] \)
\( = - \frac{1}{4} \left[ -\frac{56}{15} \right] \)
\( = \frac{56}{60} = \frac{14}{15} \)
Rationalizing the denominator is a powerful technique for simplifying integrals with square roots in the denominator.
In simple words: First, we got rid of the square roots in the bottom part of the fraction by multiplying by its conjugate. This simplified the expression greatly. Then, we integrated each part of the simplified expression using the power rule. Finally, we put in the upper and lower limits to find the definite value of the integral.

🎯 Exam Tip: When faced with an integral that has a sum or difference of square roots in the denominator, always try to rationalize the denominator first. This often leads to a much simpler integral that can be solved using standard formulas like the power rule.