Samacheer Kalvi Class 11 Maths Solutions Chapter 11 Integral Calculus Exercise 11.10

Find the integrals of the following:

Question 1.
(i) \( \frac{1}{4-x^{2}} \)
Answer:
We need to find the integral of the given expression.
\( \int \frac { dx }{ 4-x^{2} } \)
We can rewrite the denominator as \( 2^{2}-x^{2} \).
\( \int \frac { dx }{ 2^{2}-x^{2} } \)
Using the standard integration formula \( \int \frac { dx }{ a^{2}-x^{2} } = \frac { 1 }{ 2a } \log \left| \frac { a+x }{ a-x } \right| + c \), where \( a=2 \).
Substitute \( a=2 \) into the formula:
\( \frac { 1 }{ 2(2) } \log \left| \frac { 2+x }{ 2-x } \right| + c \)
Simplify the expression:
\( \frac { 1 }{ 4 } \log \left| \frac { 2+x }{ 2-x } \right| + c \)
In simple words: To solve this, we rewrite the bottom part to fit a special formula. Then we use that formula by putting in the numbers, which helps us find the final answer.

🎯 Exam Tip: Always recognize the form of the integrand (e.g., \( \frac{1}{a^2-x^2} \)) to choose the correct standard integration formula. Remember to simplify constants.

Question 1.
(ii) \( \frac{1}{25-4 x^{2}} \)
Answer:
We need to find the integral of the given expression.
\( \int \frac { 1 }{ 25-4x^{2} } dx \)
First, factor out 4 from the denominator to make the \( x^{2} \) term's coefficient 1.
\( = \frac { 1 }{ 4 } \int \frac { 1 }{ \frac { 25 }{ 4 }-x^{2} } dx \)
Now, rewrite \( \frac { 25 }{ 4 } \) as \( (\frac { 5 }{ 2 })^{2} \).
\( = \frac { 1 }{ 4 } \int \frac { dx }{ (\frac { 5 }{ 2 })^{2}-x^{2} } \)
Using the formula \( \int \frac { dx }{ a^{2}-x^{2} } = \frac { 1 }{ 2a } \log \left| \frac { a+x }{ a-x } \right| + c \), where \( a=\frac { 5 }{ 2 } \).
Substitute \( a=\frac { 5 }{ 2 } \) into the formula:
\( = \frac { 1 }{ 4 } \times \frac { 1 }{ 2(\frac { 5 }{ 2 }) } \log \left| \frac { \frac { 5 }{ 2 }+x }{ \frac { 5 }{ 2 }-x } \right| + c \)
Simplify the constants outside the log and combine the terms inside the log:
\( = \frac { 1 }{ 4 } \times \frac { 1 }{ 5 } \log \left| \frac { \frac { 5+2x }{ 2 } }{ \frac { 5-2x }{ 2 } } \right| + c \)
\( = \frac { 1 }{ 20 } \log \left| \frac { 5+2x }{ 5-2x } \right| + c \)
In simple words: First, we change the bottom part of the fraction so it fits a common math rule. Then, we use that rule with the correct numbers. This gives us the final answer, which includes a logarithm.

🎯 Exam Tip: When the coefficient of \( x^{2} \) is not 1, factor it out first to match the standard integral forms. This is a crucial first step for many integral problems.

Question 1.
(iii) \( \frac{1}{9 x^{2}-4} \)
Answer:
We need to find the integral of the given expression.
\( \int \frac { dx }{ 9x^{2}-4 } \)
First, factor out 9 from the denominator.
\( = \frac { 1 }{ 9 } \int \frac { dx }{ x^{2}-\frac { 4 }{ 9 } } \)
Rewrite \( \frac { 4 }{ 9 } \) as \( (\frac { 2 }{ 3 })^{2} \).
\( = \frac { 1 }{ 9 } \int \frac { dx }{ x^{2}-(\frac { 2 }{ 3 })^{2} } \)
Using the formula \( \int \frac { dx }{ x^{2}-a^{2} } = \frac { 1 }{ 2a } \log \left| \frac { x-a }{ x+a } \right| + c \), where \( a=\frac { 2 }{ 3 } \).
Substitute \( a=\frac { 2 }{ 3 } \) into the formula:
\( = \frac { 1 }{ 9 } \times \frac { 1 }{ 2(\frac { 2 }{ 3 }) } \log \left| \frac { x-\frac { 2 }{ 3 } }{ x+\frac { 2 }{ 3 } } \right| + c \)
Simplify the constants and combine the terms inside the log:
\( = \frac { 1 }{ 9 } \times \frac { 1 }{ \frac { 4 }{ 3 } } \log \left| \frac { \frac { 3x-2 }{ 3 } }{ \frac { 3x+2 }{ 3 } } \right| + c \)
\( = \frac { 1 }{ 9 } \times \frac { 3 }{ 4 } \log \left| \frac { 3x-2 }{ 3x+2 } \right| + c \)
\( = \frac { 1 }{ 12 } \log \left| \frac { 3x-2 }{ 3x+2 } \right| + c \)
In simple words: We change the bottom part of the fraction to match a known math rule. Then we use that rule by putting in the numbers, simplifying step by step until we get the final answer.

🎯 Exam Tip: Remember to simplify the complex fraction within the logarithm by finding a common denominator for the numerator and denominator before canceling.

Question 2.
(i) \( \frac{1}{6 x-7-x^{2}} \)
Answer:
We need to find the integral of the given expression.
\( \int \frac { dx }{ 6x-7-x^{2} } \)
First, we complete the square for the denominator \( 6x-7-x^{2} \).
\( 6x-7-x^{2} = -(x^{2}-6x+7) \)
\( = -((x-3)^{2}-9+7) \)
\( = -((x-3)^{2}-2) \)
\( = 2-(x-3)^{2} \)
So, the integral becomes:
\( \int \frac { dx }{ 2-(x-3)^{2} } \)
Let \( t = x-3 \). Then, \( dt = dx \).
Substitute \( t \) into the integral:
\( = \int \frac { dt }{ (\sqrt{2})^{2}-t^{2} } \)
Using the formula \( \int \frac { dx }{ a^{2}-x^{2} } = \frac { 1 }{ 2a } \log \left| \frac { a+x }{ a-x } \right| + c \), where \( a=\sqrt{2} \).
Substitute \( a=\sqrt{2} \) and \( t \) into the formula:
\( = \frac { 1 }{ 2\sqrt{2} } \log \left| \frac { \sqrt{2}+t }{ \sqrt{2}-t } \right| + c \)
Now, substitute back \( t=x-3 \):
\( = \frac { 1 }{ 2\sqrt{2} } \log \left| \frac { \sqrt{2}+(x-3) }{ \sqrt{2}-(x-3) } \right| + c \)
In simple words: We first change the bottom part of the fraction by completing the square. Then we replace a part of the expression with a new letter and use a special math rule for integration. Finally, we put the original expression back to get our answer.

🎯 Exam Tip: Completing the square is often the first step when the denominator or expression under the root is a quadratic function, to transform it into a standard integral form.

Question 2.
(ii) \( \frac{1}{(x+1)^{2}-25} \)
Answer:
We need to find the integral of the given expression.
\( \int \frac { dx }{ (x+1)^{2}-25 } \)
We can rewrite 25 as \( 5^{2} \).
\( = \int \frac { dx }{ (x+1)^{2}-5^{2} } \)
Let \( t = x+1 \). Then, \( dt = dx \).
Substitute \( t \) into the integral:
\( = \int \frac { dt }{ t^{2}-5^{2} } \)
Using the formula \( \int \frac { dx }{ x^{2}-a^{2} } = \frac { 1 }{ 2a } \log \left| \frac { x-a }{ x+a } \right| + c \), where \( a=5 \).
Substitute \( a=5 \) and \( t \) into the formula:
\( = \frac { 1 }{ 2(5) } \log \left| \frac { t-5 }{ t+5 } \right| + c \)
Simplify the constant:
\( = \frac { 1 }{ 10 } \log \left| \frac { t-5 }{ t+5 } \right| + c \)
Now, substitute back \( t=x+1 \):
\( = \frac { 1 }{ 10 } \log \left| \frac { (x+1)-5 }{ (x+1)+5 } \right| + c \)
Simplify the numerator and denominator inside the logarithm:
\( = \frac { 1 }{ 10 } \log \left| \frac { x-4 }{ x+6 } \right| + c \)
In simple words: We first rewrite the numbers to match a standard math rule for integrals. Then we use a substitution to simplify the integral, apply the formula, and finally replace the substituted part with the original expression to get the answer.

🎯 Exam Tip: Remember to clearly show the substitution step (e.g., let \( t = x+1 \)) and substitute back to the original variable at the end to avoid losing marks.

Question 2.
(iii) \( \frac{1}{\sqrt{x^{2}+4 x+2}} \)
Answer:
We need to find the integral of the given expression.
\( \int \frac { dx }{ \sqrt{x^{2}+4x+2} } \)
First, we complete the square for the expression under the square root, \( x^{2}+4x+2 \).
\( x^{2}+4x+2 = (x+2)^{2}-4+2 \)
\( = (x+2)^{2}-2 \)
So, the integral becomes:
\( = \int \frac { dx }{ \sqrt{(x+2)^{2}-2} } \)
Let \( t = x+2 \). Then, \( dt = dx \).
Substitute \( t \) into the integral:
\( = \int \frac { dt }{ \sqrt{t^{2}-(\sqrt{2})^{2}} } \)
Using the formula \( \int \frac { dx }{ \sqrt{x^{2}-a^{2}} } = \log \left| x+\sqrt{x^{2}-a^{2}} \right| + c \), where \( a=\sqrt{2} \).
Substitute \( a=\sqrt{2} \) and \( t \) into the formula:
\( = \log \left| t+\sqrt{t^{2}-(\sqrt{2})^{2}} \right| + c \)
Now, substitute back \( t=x+2 \):
\( = \log \left| (x+2)+\sqrt{(x+2)^{2}-(\sqrt{2})^{2}} \right| + c \)
Simplify the expression under the square root:
\( = \log \left| (x+2)+\sqrt{x^{2}+4x+4-2} \right| + c \)
\( = \log \left| (x+2)+\sqrt{x^{2}+4x+2} \right| + c \)
In simple words: We change the expression under the square root by completing the square. Then we use a substitution to make the integral simpler, apply a known formula for square root integrals, and finally put back the original expression to get our answer.

🎯 Exam Tip: Always remember that \( a^{2} \) can be \( (\sqrt{a})^{2} \). This is useful when the constant term cannot be expressed as a perfect square of an integer, as seen with \( \sqrt{2} \).

Question 3.
(i) \( \frac{1}{\sqrt{(2+x)^{2}-1}} \)
Answer:
We need to find the integral of the given expression.
\( \int \frac { dx }{ \sqrt{(2+x)^{2}-1} } \)
We can rewrite 1 as \( 1^{2} \).
\( = \int \frac { dx }{ \sqrt{(2+x)^{2}-1^{2}} } \)
Let \( t = 2+x \). Then, \( dt = dx \).
Substitute \( t \) into the integral:
\( = \int \frac { dt }{ \sqrt{t^{2}-1^{2}} } \)
Using the formula \( \int \frac { dx }{ \sqrt{x^{2}-a^{2}} } = \log \left| x+\sqrt{x^{2}-a^{2}} \right| + c \), where \( a=1 \).
Substitute \( a=1 \) and \( t \) into the formula:
\( = \log \left| t+\sqrt{t^{2}-1^{2}} \right| + c \)
Now, substitute back \( t=2+x \):
\( = \log \left| (2+x)+\sqrt{(2+x)^{2}-1} \right| + c \)
In simple words: We make a substitution to simplify the integral. Then, we use a standard formula for square root integrals. After applying the formula, we put the original expression back in place of the substituted letter to get the final answer.

🎯 Exam Tip: The term \( (2+x)^2 \) is already in a perfect square form, simplifying the initial steps. Focus on correctly identifying 'a' and applying the square root integral formula.

Question 3.
(ii) \( \frac{1}{\sqrt{x^{2}-4 x+5}} \)
Answer:
We need to find the integral of the given expression.
\( \int \frac { dx }{ \sqrt{x^{2}-4x+5} } \)
First, we complete the square for the expression under the square root, \( x^{2}-4x+5 \).
\( x^{2}-4x+5 = (x-2)^{2}-4+5 \)
\( = (x-2)^{2}+1 \)
So, the integral becomes:
\( = \int \frac { dx }{ \sqrt{(x-2)^{2}+1} } \)
Let \( t = x-2 \). Then, \( dt = dx \).
Substitute \( t \) into the integral:
\( = \int \frac { dt }{ \sqrt{t^{2}+1^{2}} } \)
Using the formula \( \int \frac { dx }{ \sqrt{x^{2}+a^{2}} } = \log \left| x+\sqrt{x^{2}+a^{2}} \right| + c \), where \( a=1 \).
Substitute \( a=1 \) and \( t \) into the formula:
\( = \log \left| t+\sqrt{t^{2}+1^{2}} \right| + c \)
Now, substitute back \( t=x-2 \):
\( = \log \left| (x-2)+\sqrt{(x-2)^{2}+1} \right| + c \)
Simplify the expression under the square root:
\( = \log \left| (x-2)+\sqrt{x^{2}-4x+4+1} \right| + c \)
\( = \log \left| (x-2)+\sqrt{x^{2}-4x+5} \right| + c \)
In simple words: We change the expression under the square root by completing the square. Then we use a substitution to make the integral simpler, apply a special formula for square root integrals, and finally put back the original expression to get our answer.

🎯 Exam Tip: When completing the square, pay close attention to the sign of the constant term after adding and subtracting. This determines whether it is \( a^{2}-x^{2} \), \( x^{2}-a^{2} \), or \( x^{2}+a^{2} \).

Question 3.
(iii) \( \frac{1}{\sqrt{9+8 x-x^{2}}} \)
Answer:
We need to find the integral of the given expression.
\( \int \frac { dx }{ \sqrt{9+8x-x^{2}} } \)
First, we complete the square for the expression under the square root, \( 9+8x-x^{2} \).
\( 9+8x-x^{2} = 9-(x^{2}-8x) \)
\( = 9-((x-4)^{2}-16) \)
\( = 9-(x-4)^{2}+16 \)
\( = 25-(x-4)^{2} \)
So, the integral becomes:
\( = \int \frac { dx }{ \sqrt{25-(x-4)^{2}} } \)
We can rewrite 25 as \( 5^{2} \).
\( = \int \frac { dx }{ \sqrt{5^{2}-(x-4)^{2}} } \)
Let \( t = x-4 \). Then, \( dt = dx \).
Substitute \( t \) into the integral:
\( = \int \frac { dt }{ \sqrt{5^{2}-t^{2}} } \)
Using the formula \( \int \frac { dx }{ \sqrt{a^{2}-x^{2}} } = \sin^{-1} \left( \frac { x }{ a } \right) + c \), where \( a=5 \).
Substitute \( a=5 \) and \( t \) into the formula:
\( = \sin^{-1} \left( \frac { t }{ 5 } \right) + c \)
Now, substitute back \( t=x-4 \):
\( = \sin^{-1} \left( \frac { x-4 }{ 5 } \right) + c \)
In simple words: We change the expression under the square root by completing the square and carefully handling the negative sign. Then we use a substitution and apply a special formula for sine inverse integrals. Finally, we put back the original expression to get our answer.

🎯 Exam Tip: When completing the square with a negative \( x^{2} \) term, factor out the negative sign first, then complete the square for the remaining quadratic expression. Be careful with the signs when reintroducing the negative factor.