Samacheer Kalvi Class 12 Business Maths Solutions Chapter 2 Integral Calculus I Exercise 2.6

Question 1. Integrate the following with respect to x. \( \frac { 2x+5 }{x^2+5x-7} \)
Answer: To integrate the given expression, we use the method of substitution. We let the denominator be \( z \), so \( z = x^2+5x-7 \). Next, we find the derivative of \( z \) with respect to \( x \), which gives \( \frac {dz}{dx} = 2x+5 \). This means \( dz = (2x+5)dx \). Notice that \( (2x+5)dx \) is exactly the numerator of our original expression. Now, we can rewrite the integral in terms of \( z \).
\( \int \frac { 2x+5 }{x^2+5x-7} dx \)
\( \implies \int \frac { 1 }{z} dz \)
\( \implies \log |z| + c \)
Substitute back \( z = x^2+5x-7 \).
\( \implies \log |x^2+5x-7| + c \)
The final step is to replace \( z \) with its original expression in terms of \( x \).
In simple words: We changed the variable from \( x \) to \( z \) to make the integral easier. We chose \( z \) so that its derivative matched the top part of the fraction. After integrating with \( z \), we put \( x \) back into the answer.

🎯 Exam Tip: Always look for an expression in the denominator (or under a root) whose derivative matches the numerator (or part of it) when using substitution. This is a common pattern for logarithmic integrals.

Question 2. Integrate \( \frac {e^{3logx} }{x^4+1} \)
Answer: First, we simplify the numerator using logarithm properties: \( e^{3\log x} = e^{\log (x^3)} = x^3 \). So the integral becomes \( \int \frac {x^3}{x^4+1} dx \). We will use the substitution method to solve this. Let \( z = x^4+1 \). Then, find the derivative of \( z \) with respect to \( x \), which is \( \frac {dz}{dx} = 4x^3 \). This means \( dz = 4x^3 dx \), or \( \frac{1}{4} dz = x^3 dx \).
Now, substitute these into the integral:
\( \int \frac {x^3}{x^4+1} dx \)
\( \implies \int \frac {1}{z} \left(\frac{1}{4} dz\right) \)
\( \implies \frac{1}{4} \int \frac {1}{z} dz \)
\( \implies \frac{1}{4} \log |z| + c \)
Finally, substitute \( z = x^4+1 \) back into the expression.
\( \implies \frac{1}{4} \log |x^4+1| + c \)
In simple words: First, we simplified the top part using a log rule. Then, we used substitution by letting the bottom part be \( z \). Since the derivative of \( z \) included \( x^3 \), it helped us solve the integral easily, which became a simple log form.

🎯 Exam Tip: Remember the logarithmic property \( e^{\log f(x)} = f(x) \) as it often helps simplify expressions before integration.

Question 3. Integrate \( \frac {e^{2x} }{e^{2x}-2} \)
Answer: To solve this integral, we use the substitution method. Let \( z = e^{2x}-2 \). Next, we find the derivative of \( z \) with respect to \( x \).
\( \frac {dz}{dx} = \frac{d}{dx} (e^{2x}-2) = e^{2x} \cdot 2 - 0 = 2e^{2x} \).
From this, we get \( dz = 2e^{2x} dx \), which means \( \frac{1}{2} dz = e^{2x} dx \).
Now, we substitute these into the integral:
\( \int \frac {e^{2x} }{e^{2x}-2} dx \)
\( \implies \int \frac {1}{z} \left(\frac{1}{2} dz\right) \)
\( \implies \frac{1}{2} \int \frac {1}{z} dz \)
\( \implies \frac{1}{2} \log |z| + c \)
Finally, substitute back \( z = e^{2x}-2 \) into the expression.
\( \implies \frac{1}{2} \log |e^{2x}-2| + c \)
In simple words: We made the integral simpler by using a new variable, \( z \). We chose \( z \) to be the bottom part of the fraction. Its derivative gave us the top part (with a number), which helped us solve the integral quickly into a log form.

🎯 Exam Tip: When dealing with integrals involving exponential functions, substitution using the exponential term as \( z \) is often very effective, especially if its derivative appears elsewhere in the expression.

Question 4. Integrate \( \frac { (logx)^3 }{x} \)
Answer: We will use the substitution method to solve this integral. Let \( z = \log x \). Next, we find the derivative of \( z \) with respect to \( x \).
\( \frac {dz}{dx} = \frac{1}{x} \).
This means \( dz = \frac{1}{x} dx \).
Now, we substitute these into the integral:
\( \int \frac { (\log x)^3 }{x} dx \)
\( \implies \int z^3 dz \)
\( \implies \frac{z^{3+1}}{3+1} + c \)
\( \implies \frac{z^4}{4} + c \)
Finally, substitute back \( z = \log x \) into the expression.
\( \implies \frac{(\log x)^4}{4} + c \)
In simple words: We used substitution by setting \( z \) equal to \( \log x \). Since the derivative of \( \log x \) is \( \frac{1}{x} \), which is present in the integral, it simplified into a simple power rule integration.

🎯 Exam Tip: When you see \( \log x \) and \( \frac{1}{x} \) together in an integral, consider substituting \( z = \log x \), as \( \frac{1}{x} \) will become part of \( dz \).

Question 5. Integrate \( \frac { 6x+7 }{\sqrt{3x^2+7x-1}} \)
Answer: To integrate this expression, we use the substitution method. Let \( z = 3x^2+7x-1 \). Then we find the derivative of \( z \) with respect to \( x \).
\( \frac {dz}{dx} = 6x+7 \).
This means \( dz = (6x+7)dx \). Notice that \( (6x+7)dx \) is exactly the numerator.
Now, we substitute these into the integral:
\( \int \frac { 6x+7 }{\sqrt{3x^2+7x-1}} dx \)
\( \implies \int \frac {dz}{\sqrt{z}} \)
\( \implies \int z^{-1/2} dz \)
Using the power rule for integration \( \int u^n du = \frac{u^{n+1}}{n+1} + C \):
\( \implies \frac{z^{-1/2+1}}{-1/2+1} + c \)
\( \implies \frac{z^{1/2}}{1/2} + c \)
\( \implies 2z^{1/2} + c \)
\( \implies 2\sqrt{z} + c \)
Finally, substitute back \( z = 3x^2+7x-1 \) into the expression.
\( \implies 2\sqrt{3x^2+7x-1} + c \)
In simple words: We used substitution for the expression inside the square root. The derivative of this part matched the numerator. This changed the integral into a simpler form with \( z \), which we then solved using the power rule for roots.

🎯 Exam Tip: For integrals involving a square root in the denominator, if the derivative of the term inside the root is in the numerator, substitute \( z \) for the expression inside the root to simplify it to \( \int z^{-1/2} dz \).

Question 6. Integrate \( (4x + 2) \sqrt {x^2+x+1} \)
Answer: We can rewrite the expression as \( 2(2x+1)\sqrt{x^2+x+1} \).
Now, we will use the substitution method. Let \( z = x^2+x+1 \).
Then, find the derivative of \( z \) with respect to \( x \).
\( \frac {dz}{dx} = 2x+1 \).
This means \( dz = (2x+1)dx \).
Substitute these into the integral:
\( \int 2(2x+1)\sqrt{x^2+x+1} dx \)
\( \implies 2 \int \sqrt{z} dz \)
\( \implies 2 \int z^{1/2} dz \)
Using the power rule for integration \( \int u^n du = \frac{u^{n+1}}{n+1} + C \):
\( \implies 2 \left( \frac{z^{1/2+1}}{1/2+1} \right) + c \)
\( \implies 2 \left( \frac{z^{3/2}}{3/2} \right) + c \)
\( \implies 2 \cdot \frac{2}{3} z^{3/2} + c \)
\( \implies \frac{4}{3} z^{3/2} + c \)
Finally, substitute back \( z = x^2+x+1 \) into the expression.
\( \implies \frac{4}{3} (x^2+x+1)^{3/2} + c \)
In simple words: We first pulled out a constant and then used substitution for the expression under the square root. The derivative of that expression matched the remaining part of the integral. This simplified the integral, allowing us to use the basic power rule for integration.

🎯 Exam Tip: Always factor out any constants that are common to parts of the integrand. This can simplify the derivative match for substitution.

Question 7. Integrate \( x^8 (1 + x^9)^5 \)
Answer: We will use the substitution method for this integral. Let \( z = 1+x^9 \).
Next, find the derivative of \( z \) with respect to \( x \).
\( \frac {dz}{dx} = 9x^8 \).
This means \( dz = 9x^8 dx \), or \( \frac{1}{9} dz = x^8 dx \).
Now, substitute these into the integral:
\( \int x^8 (1+x^9)^5 dx \)
\( \implies \int z^5 \left(\frac{1}{9} dz\right) \)
\( \implies \frac{1}{9} \int z^5 dz \)
Using the power rule for integration \( \int u^n du = \frac{u^{n+1}}{n+1} + C \):
\( \implies \frac{1}{9} \left( \frac{z^{5+1}}{5+1} \right) + c \)
\( \implies \frac{1}{9} \left( \frac{z^6}{6} \right) + c \)
\( \implies \frac{1}{54} z^6 + c \)
Finally, substitute back \( z = 1+x^9 \) into the expression.
\( \implies \frac{1}{54} (1+x^9)^6 + c \)
In simple words: We used a simple substitution by letting \( z \) be the term inside the parenthesis. The derivative of \( z \) was \( 9x^8 \), which helped us easily replace \( x^8 dx \) in the integral. This turned it into a simple power rule problem.

🎯 Exam Tip: When you have a term raised to a power and another term that is its derivative (or a constant multiple of it), substitution is often the key. Identify the inner function for \( z \).

Question 8. Integrate \( \frac {x^{e-1}+e^{x-1} }{x^e+e^x} \)
Answer: We will use the substitution method for this integral. Let \( z = x^e+e^x \).
Next, find the derivative of \( z \) with respect to \( x \).
\( \frac {dz}{dx} = \frac{d}{dx}(x^e) + \frac{d}{dx}(e^x) \).
The derivative of \( x^e \) is \( e \cdot x^{e-1} \) (using power rule \( \frac{d}{dx} x^n = nx^{n-1} \)).
The derivative of \( e^x \) is \( e^x \).
So, \( \frac {dz}{dx} = e x^{e-1} + e^x \).
We can factor out \( e \) from the first term: \( \frac {dz}{dx} = e (x^{e-1} + e^{x-1}) \).
This means \( dz = e (x^{e-1} + e^{x-1}) dx \), or \( \frac{1}{e} dz = (x^{e-1} + e^{x-1}) dx \).
Now, substitute these into the integral:
\( \int \frac {x^{e-1}+e^{x-1} }{x^e+e^x} dx \)
\( \implies \int \frac{1}{z} \left(\frac{1}{e} dz\right) \)
\( \implies \frac{1}{e} \int \frac{1}{z} dz \)
\( \implies \frac{1}{e} \log |z| + c \)
Finally, substitute back \( z = x^e+e^x \) into the expression.
\( \implies \frac{1}{e} \log |x^e+e^x| + c \)
In simple words: We chose \( z \) to be the entire bottom part of the fraction. When we found the derivative of \( z \), we noticed it matched the top part (with a constant factor). This allowed us to change the integral into a simple log form, which is easy to solve.

🎯 Exam Tip: Remember that \( \frac{d}{dx} x^e = e x^{e-1} \) (using the power rule) and \( \frac{d}{dx} e^x = e^x \). Recognizing these derivatives is crucial for this type of substitution.

Question 9. Integrate \( \frac { 1 }{x \log x} \)
Answer: To integrate this, we use the substitution method. Let \( z = \log x \).
Next, we find the derivative of \( z \) with respect to \( x \).
\( \frac {dz}{dx} = \frac{1}{x} \).
This means \( dz = \frac{1}{x} dx \).
Now, we substitute these into the integral:
\( \int \frac { 1 }{x \log x} dx \)
We can rewrite this as \( \int \frac {1}{\log x} \cdot \frac{1}{x} dx \).
\( \implies \int \frac {1}{z} dz \)
\( \implies \log |z| + c \)
Finally, substitute back \( z = \log x \) into the expression.
\( \implies \log |\log x| + c \)
In simple words: We made the integral easier by setting \( z \) to \( \log x \). Since the derivative of \( \log x \) is \( \frac{1}{x} \), which was also in the problem, the integral became a simple log integration problem.

🎯 Exam Tip: Whenever you see a \( \log x \) term and a \( \frac{1}{x} \) term in an integral, consider substitution with \( z = \log x \). This often simplifies the integral significantly.

Question 10. Integrate \( \frac { x }{2x^4-3x^2-2} \)
Answer: To solve this integral, we first use substitution. Let \( t = x^2 \). Then \( \frac{dt}{dx} = 2x \), which means \( dx = \frac{dt}{2x} \).
The integral becomes:
\( \int \frac { x }{2t^2-3t-2} \cdot \frac{dt}{2x} \)
\( \implies \int \frac { 1 }{2(2t^2-3t-2)} dt \)
We factor the denominator: \( 2t^2-3t-2 = (2t+1)(t-2) \).
So, we have \( \frac{1}{2} \int \frac{1}{(2t+1)(t-2)} dt \).
Now, we use partial fractions for \( \frac{1}{(2t+1)(t-2)} \).
Let \( \frac{1}{(2t+1)(t-2)} = \frac{A}{2t+1} + \frac{B}{t-2} \).
Multiplying by \( (2t+1)(t-2) \): \( 1 = A(t-2) + B(2t+1) \).
Set \( t = 2 \): \( 1 = A(2-2) + B(2(2)+1) \implies 1 = 5B \implies B = \frac{1}{5} \).
Set \( t = -\frac{1}{2} \): \( 1 = A(-\frac{1}{2}-2) + B(0) \implies 1 = A(-\frac{5}{2}) \implies A = -\frac{2}{5} \).
So, \( \frac{1}{(2t+1)(t-2)} = \frac{-2/5}{2t+1} + \frac{1/5}{t-2} \).
Substitute this back into the integral:
\( \frac{1}{2} \int \left( \frac{-2/5}{2t+1} + \frac{1/5}{t-2} \right) dt \)
\( \implies \frac{1}{2} \left[ -\frac{2}{5} \int \frac{1}{2t+1} dt + \frac{1}{5} \int \frac{1}{t-2} dt \right] \)
\( \implies \frac{1}{2} \left[ -\frac{2}{5} \cdot \frac{1}{2} \log|2t+1| + \frac{1}{5} \log|t-2| \right] + c \)
\( \implies \frac{1}{2} \left[ -\frac{1}{5} \log|2t+1| + \frac{1}{5} \log|t-2| \right] + c \)
\( \implies \frac{1}{10} [ \log|t-2| - \log|2t+1| ] + c \)
\( \implies \frac{1}{10} \log \left| \frac{t-2}{2t+1} \right| + c \)
Finally, substitute back \( t = x^2 \):
\( \implies \frac{1}{10} \log \left| \frac{x^2-2}{2x^2+1} \right| + c \)
In simple words: We first used substitution to change \( x^2 \) to \( t \), which simplified the denominator. Then, we broke the fraction into simpler parts using partial fractions. After integrating each part separately (which gave log terms), we put \( x^2 \) back in place of \( t \) to get the final answer.

🎯 Exam Tip: For integrals with polynomial denominators that can be factored, partial fraction decomposition is a powerful technique. Remember to perform an initial substitution if the power of \( x \) in the numerator is lower than the main power in the denominator, and the variables are related (like \( x \) and \( x^2 \) or \( x^3 \)).

Question 11. Integrate \( e^x (1 + x) \log (x e^x) \)
Answer: We use the substitution method for this integral. Let \( z = x e^x \).
Next, we find the derivative of \( z \) with respect to \( x \) using the product rule \( (uv)' = u'v + uv' \).
\( \frac {dz}{dx} = \frac{d}{dx}(x) \cdot e^x + x \cdot \frac{d}{dx}(e^x) \)
\( \implies \frac {dz}{dx} = 1 \cdot e^x + x \cdot e^x \)
\( \implies \frac {dz}{dx} = e^x(1+x) \)
So, \( dz = e^x(1+x)dx \).
Now, substitute these into the integral:
\( \int e^x (1 + x) \log (x e^x) dx \)
\( \implies \int \log(z) dz \)
We know the integral of \( \log(u) du \) is \( u(\log u - 1) + C \).
So, \( \int \log(z) dz = z(\log z - 1) + c \).
Finally, substitute back \( z = x e^x \) into the expression.
\( \implies x e^x (\log (x e^x) - 1) + c \)
In simple words: We picked \( z \) as \( x e^x \). When we found the derivative of \( z \), it exactly matched the other part of the integral, \( e^x(1+x)dx \). This changed the problem into a simple integral of \( \log z \), which we solved using a known formula.

🎯 Exam Tip: The derivative of \( x e^x \) is a common pattern to recognize, as it is \( e^x(1+x) \). This recognition is key for substitution problems involving this product.

Question 12. Integrate \( \frac { 1 }{x(x^2+1)} \)
Answer: We use the method of partial fractions to solve this integral. The denominator has a linear factor \( x \) and an irreducible quadratic factor \( x^2+1 \).
So, we can write the expression as:
\( \frac{1}{x(x^2+1)} = \frac{A}{x} + \frac{Bx+C}{x^2+1} \)
Multiply both sides by \( x(x^2+1) \):
\( 1 = A(x^2+1) + (Bx+C)x \)
\( 1 = Ax^2+A + Bx^2+Cx \)
\( 1 = (A+B)x^2 + Cx + A \)
Comparing the coefficients of the powers of \( x \):
Coefficient of \( x^2 \): \( A+B = 0 \)
Coefficient of \( x \): \( C = 0 \)
Constant term: \( A = 1 \)
From \( A=1 \) and \( A+B=0 \), we get \( 1+B=0 \implies B = -1 \).
So the partial fraction decomposition is:
\( \frac{1}{x(x^2+1)} = \frac{1}{x} + \frac{-x+0}{x^2+1} = \frac{1}{x} - \frac{x}{x^2+1} \)
Now, we integrate each term:
\( \int \left( \frac{1}{x} - \frac{x}{x^2+1} \right) dx \)
\( \implies \int \frac{1}{x} dx - \int \frac{x}{x^2+1} dx \)
The first integral is \( \log |x| \).
For the second integral, let \( u = x^2+1 \), then \( du = 2x dx \), so \( x dx = \frac{1}{2} du \).
\( \int \frac{x}{x^2+1} dx = \int \frac{1}{u} \cdot \frac{1}{2} du = \frac{1}{2} \log |u| = \frac{1}{2} \log |x^2+1| \).
Combine the results:
\( \implies \log |x| - \frac{1}{2} \log |x^2+1| + c \)
Using logarithm properties, \( \log a - \log b = \log (a/b) \) and \( k \log a = \log a^k \):
\( \implies \log |x| - \log |(x^2+1)^{1/2}| + c \)
\( \implies \log \left| \frac{x}{\sqrt{x^2+1}} \right| + c \)
In simple words: We broke the complicated fraction into two simpler ones using a method called partial fractions. We found the correct numbers for these new fractions. Then, we integrated each simple fraction separately. One part was a basic log integral, and the other also turned into a log integral after a small substitution.

🎯 Exam Tip: When using partial fractions, remember to use the form \( \frac{Bx+C}{x^2+1} \) for irreducible quadratic factors in the denominator, not just \( \frac{B}{x^2+1} \).

Question 13. Integrate \( e^x \left[ \frac { 1 }{x^2} - \frac { 2 }{x^3} \right] \)
Answer: This integral is of the form \( \int e^x [f(x) + f'(x)] dx \), which integrates to \( e^x f(x) + C \).
We need to identify \( f(x) \) and \( f'(x) \) in the given expression.
Let \( f(x) = \frac{1}{x^2} \).
Now, let's find the derivative of \( f(x) \).
\( f(x) = x^{-2} \)
\( f'(x) = -2x^{-2-1} = -2x^{-3} = -\frac{2}{x^3} \).
The given expression is \( e^x \left[ \frac { 1 }{x^2} - \frac { 2 }{x^3} \right] \), which is \( e^x [f(x) + f'(x)] \).
Therefore, the integral is:
\( \int e^x \left[ \frac { 1 }{x^2} - \frac { 2 }{x^3} \right] dx = e^x f(x) + c \)
\( \implies e^x \left( \frac{1}{x^2} \right) + c \)
In simple words: This problem uses a special rule where if an integral looks like \( e^x \) multiplied by a function plus its derivative, the answer is simply \( e^x \) times that function. We found that \( \frac{1}{x^2} \) was the function, and \( -\frac{2}{x^3} \) was its derivative, so we applied the rule directly.

🎯 Exam Tip: Always be on the lookout for the \( \int e^x [f(x) + f'(x)] dx \) form. It's a direct formula that can save a lot of time in integration problems.

Question 14. Integrate \( e^x \left[ \frac {x-1 }{(x+1)^3} \right] \)
Answer: This integral can also be solved using the form \( \int e^x [f(x) + f'(x)] dx = e^x f(x) + C \).
We need to manipulate the term \( \frac{x-1}{(x+1)^3} \) to fit the \( f(x) + f'(x) \) pattern.
We can rewrite the numerator \( x-1 \) as \( (x+1)-2 \).
So, \( \frac{x-1}{(x+1)^3} = \frac{(x+1)-2}{(x+1)^3} \)
Split this into two fractions:
\( \implies \frac{x+1}{(x+1)^3} - \frac{2}{(x+1)^3} \)
\( \implies \frac{1}{(x+1)^2} - \frac{2}{(x+1)^3} \)
Now, let \( f(x) = \frac{1}{(x+1)^2} = (x+1)^{-2} \).
Let's find the derivative of \( f(x) \):
\( f'(x) = -2(x+1)^{-2-1} \cdot \frac{d}{dx}(x+1) \) (using chain rule)
\( f'(x) = -2(x+1)^{-3} \cdot 1 \)
\( f'(x) = -\frac{2}{(x+1)^3} \)
So, the expression is indeed in the form \( e^x [f(x) + f'(x)] \).
Therefore, the integral is:
\( \int e^x \left[ \frac {x-1 }{(x+1)^3} \right] dx = e^x f(x) + c \)
\( \implies e^x \left( \frac{1}{(x+1)^2} \right) + c \)
In simple words: We changed the fraction inside the integral to make it fit a special rule. We rewrote the top part to create two separate fractions, one being a function and the other its derivative. With this, the answer became simply \( e^x \) multiplied by the first function.

🎯 Exam Tip: When faced with fractions inside the \( \int e^x [...] dx \) form, try to algebraically manipulate the fraction to isolate a function \( f(x) \) and its derivative \( f'(x) \). Often, adding and subtracting a term in the numerator is helpful.

Question 15. Integrate \( e^{3x} \left[ \frac { 3x-1 }{9x^2} \right] \)
Answer: This integral is of the form \( \int e^{ax} [a f(x) + f'(x)] dx \), which integrates to \( e^{ax} f(x) + C \).
In this case, \( a=3 \). We need to express \( \frac{3x-1}{9x^2} \) in the form \( 3f(x) + f'(x) \).
First, split the fraction:
\( \frac{3x-1}{9x^2} = \frac{3x}{9x^2} - \frac{1}{9x^2} = \frac{1}{3x} - \frac{1}{9x^2} \)
Now, we need to find \( f(x) \) such that \( 3f(x) + f'(x) = \frac{1}{3x} - \frac{1}{9x^2} \).
Let's try \( f(x) = \frac{1}{9x} \).
Then, \( a f(x) = 3 \cdot \frac{1}{9x} = \frac{1}{3x} \). This matches the first term.
Next, find \( f'(x) \):
\( f(x) = \frac{1}{9} x^{-1} \)
\( f'(x) = \frac{1}{9} (-1) x^{-2} = -\frac{1}{9x^2} \). This matches the second term.
So, we have successfully identified \( f(x) = \frac{1}{9x} \).
Therefore, the integral is:
\( \int e^{3x} \left[ \frac { 3x-1 }{9x^2} \right] dx = e^{3x} f(x) + c \)
\( \implies e^{3x} \left( \frac{1}{9x} \right) + c \)
\( \implies \frac{e^{3x}}{9x} + c \)
In simple words: This integral uses a special rule for \( e^{ax} \) type problems. We split the fraction into two parts and identified a function \( f(x) \) whose derivative, when combined with \( a \) times the function itself, matched the entire fractional part. Once found, the answer is just \( e^{ax} \) multiplied by \( f(x) \).

🎯 Exam Tip: When integrating expressions of the form \( \int e^{ax} \cdot g(x) dx \), always check if \( g(x) \) can be written as \( a f(x) + f'(x) \). If so, the solution is simply \( e^{ax} f(x) + C \).