Download Exercise 7.4 PDF
- Ex. 7.1
- Ex. 7.2
- Ex. 7.2
- Ex. 7.3
- Ex. 7.4
- Ex. 7.5
- Ex. 7.6
- Ex. 7.7
- Ex. 7.8
- Ex. 7.9
- Ex. 7.10
- Ex. 7.11
- Miscellaneous Exercise
Integrals Exercise 7.4 Class 12 Mathematics Solutions
Detailed step-by-step solutions for CBSE Class 12 Mathematics, Chapter 7 Integrals, Exercise 7.4. This exercise deals with integrals of some particular functions involving completing the square and standard integration formulas.
Question 1: Integrate the function \(\frac{3x^2}{x^6+1}\)
Let \(I = \int \frac{3x^2}{x^6+1} dx\)
We can write \(x^6 = (x^3)^2\).
\(I = \int \frac{3x^2}{(x^3)^2+1} dx\)
We can write \(x^6 = (x^3)^2\).
\(I = \int \frac{3x^2}{(x^3)^2+1} dx\)
Substitute \(x^3 = t\).
Differentiating both sides: \(3x^2 dx = dt\).
Differentiating both sides: \(3x^2 dx = dt\).
Now substitute these values in the integral:
\(I = \int \frac{dt}{t^2+1}\)
\(I = \int \frac{dt}{t^2+1}\)
We know that \(\int \frac{dx}{x^2+1} = \tan^{-1}x + C\).
Therefore, \(I = \tan^{-1}t + C\)
Therefore, \(I = \tan^{-1}t + C\)
Putting back \(t = x^3\):
\(I = \tan^{-1}(x^3) + C\)
\(I = \tan^{-1}(x^3) + C\)
Question 2: Integrate the function \(\frac{1}{\sqrt{1+4x^2}}\)
Let \(I = \int \frac{1}{\sqrt{1+4x^2}} dx = \int \frac{1}{\sqrt{1+(2x)^2}} dx\)
Put \(2x = t\).
\(\Rightarrow 2dx = dt \Rightarrow dx = \frac{dt}{2}\).
\(\Rightarrow 2dx = dt \Rightarrow dx = \frac{dt}{2}\).
\(I = \frac{1}{2} \int \frac{dt}{\sqrt{1+t^2}}\)
Using the formula \(\int \frac{dx}{\sqrt{x^2+a^2}} = \log|x+\sqrt{x^2+a^2}| + C\):
Using the formula \(\int \frac{dx}{\sqrt{x^2+a^2}} = \log|x+\sqrt{x^2+a^2}| + C\):
\(I = \frac{1}{2} \log|t + \sqrt{t^2+1}| + C\)
Substitute \(t = 2x\):
\(I = \frac{1}{2} \log|2x + \sqrt{4x^2+1}| + C\)
\(I = \frac{1}{2} \log|2x + \sqrt{4x^2+1}| + C\)
Question 3: Integrate the function \(\frac{1}{\sqrt{(2-x)^2+1}}\)
Let \(2-x = t\).
\(\Rightarrow -dx = dt \Rightarrow dx = -dt\).
\(\Rightarrow -dx = dt \Rightarrow dx = -dt\).
Integral becomes:
\(I = \int \frac{-dt}{\sqrt{t^2+1}} = - \int \frac{dt}{\sqrt{t^2+1}}\)
\(I = \int \frac{-dt}{\sqrt{t^2+1}} = - \int \frac{dt}{\sqrt{t^2+1}}\)
Using standard formula:
\(I = - \log|t + \sqrt{t^2+1}| + C\)
\(I = - \log|t + \sqrt{t^2+1}| + C\)
Substitute \(t = 2-x\):
\(I = - \log|(2-x) + \sqrt{(2-x)^2+1}| + C\)
Or
\(I = \log \left| \frac{1}{(2-x) + \sqrt{x^2-4x+5}} \right| + C\)
\(I = - \log|(2-x) + \sqrt{(2-x)^2+1}| + C\)
Or
\(I = \log \left| \frac{1}{(2-x) + \sqrt{x^2-4x+5}} \right| + C\)
Question 4: Integrate the function \(\frac{1}{\sqrt{9-25x^2}}\)
Let \(I = \int \frac{1}{\sqrt{9-25x^2}} dx = \int \frac{1}{\sqrt{3^2-(5x)^2}} dx\)
Put \(5x = t \Rightarrow 5dx = dt \Rightarrow dx = \frac{dt}{5}\).
\(I = \frac{1}{5} \int \frac{dt}{\sqrt{3^2-t^2}}\)
Using \(\int \frac{dx}{\sqrt{a^2-x^2}} = \sin^{-1}(\frac{x}{a}) + C\):
\(I = \frac{1}{5} \sin^{-1}(\frac{t}{3}) + C\)
\(I = \frac{1}{5} \sin^{-1}(\frac{t}{3}) + C\)
\(I = \frac{1}{5} \sin^{-1}(\frac{5x}{3}) + C\)
Question 5: Integrate the function \(\frac{3x}{1+2x^4}\)
Write \(2x^4\) as \((\sqrt{2}x^2)^2\).
Let \(I = \int \frac{3x}{1+(\sqrt{2}x^2)^2} dx\)
Let \(I = \int \frac{3x}{1+(\sqrt{2}x^2)^2} dx\)
Put \(\sqrt{2}x^2 = t\).
\(\Rightarrow 2\sqrt{2}x dx = dt \Rightarrow x dx = \frac{dt}{2\sqrt{2}}\).
\(\Rightarrow 2\sqrt{2}x dx = dt \Rightarrow x dx = \frac{dt}{2\sqrt{2}}\).
\(I = \frac{3}{2\sqrt{2}} \int \frac{dt}{1+t^2}\)
\(I = \frac{3}{2\sqrt{2}} \tan^{-1}t + C\)
\(I = \frac{3}{2\sqrt{2}} \tan^{-1}(\sqrt{2}x^2) + C\)
\(I = \frac{3}{2\sqrt{2}} \tan^{-1}(\sqrt{2}x^2) + C\)
Question 6: Integrate the function \(\frac{x^2}{1-x^6}\)
Let \(x^3 = t \Rightarrow 3x^2 dx = dt \Rightarrow x^2 dx = \frac{dt}{3}\).
\(I = \int \frac{x^2}{1-(x^3)^2} dx = \frac{1}{3} \int \frac{dt}{1-t^2}\)
Using \(\int \frac{dx}{a^2-x^2} = \frac{1}{2a} \log|\frac{a+x}{a-x}| + C\):
\(I = \frac{1}{3} \cdot \frac{1}{2(1)} \log|\frac{1+t}{1-t}| + C\)
\(I = \frac{1}{3} \cdot \frac{1}{2(1)} \log|\frac{1+t}{1-t}| + C\)
\(I = \frac{1}{6} \log|\frac{1+x^3}{1-x^3}| + C\)
Question 7: Integrate the function \(\frac{x-1}{\sqrt{x^2-1}}\)
Separate the integral into two parts:
\(I = \int \frac{x}{\sqrt{x^2-1}} dx - \int \frac{1}{\sqrt{x^2-1}} dx\)
\(I = \int \frac{x}{\sqrt{x^2-1}} dx - \int \frac{1}{\sqrt{x^2-1}} dx\)
For the first part \(\int \frac{x}{\sqrt{x^2-1}} dx\):
Put \(x^2-1 = t \Rightarrow 2x dx = dt \Rightarrow x dx = dt/2\).
\(I_1 = \frac{1}{2} \int t^{-1/2} dt = \frac{1}{2} \cdot 2t^{1/2} = \sqrt{t} = \sqrt{x^2-1}\).
Put \(x^2-1 = t \Rightarrow 2x dx = dt \Rightarrow x dx = dt/2\).
\(I_1 = \frac{1}{2} \int t^{-1/2} dt = \frac{1}{2} \cdot 2t^{1/2} = \sqrt{t} = \sqrt{x^2-1}\).
For the second part \(\int \frac{1}{\sqrt{x^2-1}} dx\):
Using formula \(\log|x+\sqrt{x^2-1}|\).
Using formula \(\log|x+\sqrt{x^2-1}|\).
\(I = \sqrt{x^2-1} - \log|x+\sqrt{x^2-1}| + C\)
Question 8: Integrate the function \(\frac{x^2}{\sqrt{x^6+a^6}}\)
Let \(x^3 = t \Rightarrow 3x^2 dx = dt \Rightarrow x^2 dx = \frac{dt}{3}\).
Denominator becomes \(\sqrt{t^2 + (a^3)^2}\).
Denominator becomes \(\sqrt{t^2 + (a^3)^2}\).
\(I = \frac{1}{3} \int \frac{dt}{\sqrt{t^2+(a^3)^2}}\)
Using \(\int \frac{dx}{\sqrt{x^2+A^2}} = \log|x+\sqrt{x^2+A^2}|\):
\(I = \frac{1}{3} \log|t + \sqrt{t^2+a^6}| + C\)
\(I = \frac{1}{3} \log|x^3 + \sqrt{x^6+a^6}| + C\)
\(I = \frac{1}{3} \log|x^3 + \sqrt{x^6+a^6}| + C\)
Question 9: Integrate the function \(\frac{\sec^2 x}{\sqrt{\tan^2 x + 4}}\)
Let \(\tan x = t \Rightarrow \sec^2 x dx = dt\).
\(I = \int \frac{dt}{\sqrt{t^2+2^2}}\)
\(I = \log|t + \sqrt{t^2+4}| + C\)
\(I = \log|\tan x + \sqrt{\tan^2 x + 4}| + C\)
\(I = \log|\tan x + \sqrt{\tan^2 x + 4}| + C\)
Question 10: Integrate the function \(\frac{1}{\sqrt{x^2+2x+2}}\)
Complete the square in the denominator:
\(x^2+2x+2 = (x^2+2x+1) + 1 = (x+1)^2 + 1^2\).
\(x^2+2x+2 = (x^2+2x+1) + 1 = (x+1)^2 + 1^2\).
\(I = \int \frac{dx}{\sqrt{(x+1)^2+1}}\)
Put \(x+1 = t \Rightarrow dx = dt\).
\(I = \int \frac{dt}{\sqrt{t^2+1}}\)
\(I = \int \frac{dt}{\sqrt{t^2+1}}\)
\(I = \log|t+\sqrt{t^2+1}| + C\)
\(I = \log|x+1+\sqrt{x^2+2x+2}| + C\)
\(I = \log|x+1+\sqrt{x^2+2x+2}| + C\)
Question 11: Integrate the function \(\frac{1}{9x^2+6x+5}\)
Factor out 9 from the denominator:
\(9x^2+6x+5 = 9(x^2 + \frac{2}{3}x + \frac{5}{9})\).
Complete the square: \(x^2 + \frac{2}{3}x + (\frac{1}{3})^2 - (\frac{1}{3})^2 + \frac{5}{9}\)
\(= (x+\frac{1}{3})^2 + \frac{4}{9} = (x+\frac{1}{3})^2 + (\frac{2}{3})^2\).
\(9x^2+6x+5 = 9(x^2 + \frac{2}{3}x + \frac{5}{9})\).
Complete the square: \(x^2 + \frac{2}{3}x + (\frac{1}{3})^2 - (\frac{1}{3})^2 + \frac{5}{9}\)
\(= (x+\frac{1}{3})^2 + \frac{4}{9} = (x+\frac{1}{3})^2 + (\frac{2}{3})^2\).
\(I = \frac{1}{9} \int \frac{dx}{(x+\frac{1}{3})^2 + (\frac{2}{3})^2}\)
Let \(x+\frac{1}{3} = t \Rightarrow dx = dt\).
Using \(\int \frac{dx}{x^2+a^2} = \frac{1}{a}\tan^{-1}(\frac{x}{a})\):
\(I = \frac{1}{9} \cdot \frac{1}{2/3} \tan^{-1}\left(\frac{t}{2/3}\right) + C\)
\(I = \frac{1}{9} \cdot \frac{1}{2/3} \tan^{-1}\left(\frac{t}{2/3}\right) + C\)
\(I = \frac{1}{6} \tan^{-1}\left(\frac{3x+1}{2}\right) + C\)
Question 12: Integrate the function \(\frac{1}{\sqrt{7-6x-x^2}}\)
Consider the term inside the root: \(7 - (x^2+6x)\).
Complete the square: \(7 - (x^2+6x+9-9) = 7 - ((x+3)^2 - 9) = 16 - (x+3)^2 = 4^2 - (x+3)^2\).
Complete the square: \(7 - (x^2+6x+9-9) = 7 - ((x+3)^2 - 9) = 16 - (x+3)^2 = 4^2 - (x+3)^2\).
\(I = \int \frac{dx}{\sqrt{4^2 - (x+3)^2}}\)
Using \(\int \frac{dx}{\sqrt{a^2-x^2}} = \sin^{-1}(\frac{x}{a})\):
\(I = \sin^{-1}\left(\frac{x+3}{4}\right) + C\)
Question 13: Integrate the function \(\frac{1}{\sqrt{(x-1)(x-2)}}\)
Expand the denominator: \((x-1)(x-2) = x^2-3x+2\).
Complete the square: \(x^2 - 3x + (\frac{3}{2})^2 - (\frac{3}{2})^2 + 2\)
\(= (x-\frac{3}{2})^2 - \frac{9}{4} + \frac{8}{4} = (x-\frac{3}{2})^2 - (\frac{1}{2})^2\).
Complete the square: \(x^2 - 3x + (\frac{3}{2})^2 - (\frac{3}{2})^2 + 2\)
\(= (x-\frac{3}{2})^2 - \frac{9}{4} + \frac{8}{4} = (x-\frac{3}{2})^2 - (\frac{1}{2})^2\).
\(I = \int \frac{dx}{\sqrt{(x-\frac{3}{2})^2 - (\frac{1}{2})^2}}\)
\(I = \log\left| (x-\frac{3}{2}) + \sqrt{x^2-3x+2} \right| + C\)
Question 14: Integrate the function \(\frac{1}{\sqrt{8+3x-x^2}}\)
Inside the root: \(- (x^2-3x-8) = - (x^2-3x+\frac{9}{4}-\frac{9}{4}-8)\)
\(= - [(x-\frac{3}{2})^2 - \frac{41}{4}] = \frac{41}{4} - (x-\frac{3}{2})^2 = (\frac{\sqrt{41}}{2})^2 - (x-\frac{3}{2})^2\).
\(= - [(x-\frac{3}{2})^2 - \frac{41}{4}] = \frac{41}{4} - (x-\frac{3}{2})^2 = (\frac{\sqrt{41}}{2})^2 - (x-\frac{3}{2})^2\).
\(I = \int \frac{dx}{\sqrt{(\frac{\sqrt{41}}{2})^2 - (x-\frac{3}{2})^2}}\)
\(I = \sin^{-1}\left( \frac{x-3/2}{\sqrt{41}/2} \right) + C\)
\(I = \sin^{-1}\left( \frac{2x-3}{\sqrt{41}} \right) + C\)
\(I = \sin^{-1}\left( \frac{2x-3}{\sqrt{41}} \right) + C\)
Question 15: Integrate the function \(\frac{1}{\sqrt{(x-a)(x-b)}}\)
Denominator: \(\sqrt{x^2-(a+b)x+ab}\).
Complete square: \((x-\frac{a+b}{2})^2 - (\frac{a+b}{2})^2 + ab\)
The constant term becomes \(-\frac{(a+b)^2 - 4ab}{4} = -\frac{(a-b)^2}{4}\).
Complete square: \((x-\frac{a+b}{2})^2 - (\frac{a+b}{2})^2 + ab\)
The constant term becomes \(-\frac{(a+b)^2 - 4ab}{4} = -\frac{(a-b)^2}{4}\).
So, denominator is \(\sqrt{(x-\frac{a+b}{2})^2 - (\frac{a-b}{2})^2}\).
\(I = \log \left| x - \frac{a+b}{2} + \sqrt{(x-a)(x-b)} \right| + C\)
Question 16: Integrate the function \(\frac{4x+1}{\sqrt{2x^2+x-3}}\)
Notice that \(\frac{d}{dx}(2x^2+x-3) = 4x+1\).
Put \(2x^2+x-3 = t \Rightarrow (4x+1)dx = dt\).
\(I = \int \frac{dt}{\sqrt{t}} = \int t^{-1/2} dt = 2\sqrt{t} + C\).
\(I = 2\sqrt{2x^2+x-3} + C\)
Question 17: Integrate the function \(\frac{x+2}{\sqrt{x^2-1}}\)
Split the integral: \(I = \int \frac{x}{\sqrt{x^2-1}} dx + \int \frac{2}{\sqrt{x^2-1}} dx\).
Part 1: Let \(x^2-1=t \Rightarrow 2xdx=dt\). Integral is \(\frac{1}{2}\int t^{-1/2}dt = \sqrt{t} = \sqrt{x^2-1}\).
Part 2: \(2 \int \frac{dx}{\sqrt{x^2-1}} = 2 \log|x+\sqrt{x^2-1}|\).
\(I = \sqrt{x^2-1} + 2\log|x+\sqrt{x^2-1}| + C\)
Question 18: Integrate the function \(\frac{5x-2}{1+2x+3x^2}\)
Let \(5x-2 = A\frac{d}{dx}(1+2x+3x^2) + B\)
\(5x-2 = A(2+6x) + B\)
Comparing coefficients:
\(6A = 5 \Rightarrow A = 5/6\)
\(2A + B = -2 \Rightarrow 2(5/6) + B = -2 \Rightarrow B = -11/3\).
\(5x-2 = A(2+6x) + B\)
Comparing coefficients:
\(6A = 5 \Rightarrow A = 5/6\)
\(2A + B = -2 \Rightarrow 2(5/6) + B = -2 \Rightarrow B = -11/3\).
\(I = \frac{5}{6} \int \frac{2+6x}{1+2x+3x^2} dx - \frac{11}{3} \int \frac{1}{3x^2+2x+1} dx\)
First part integrates to \(\frac{5}{6} \log|1+2x+3x^2|\).
Second part: \(3x^2+2x+1 = 3(x^2+\frac{2}{3}x+\frac{1}{3}) = 3[(x+\frac{1}{3})^2 + (\frac{\sqrt{2}}{3})^2]\).
Integral is \(\frac{1}{3} \cdot \frac{1}{\sqrt{2}/3} \tan^{-1}(\frac{3x+1}{\sqrt{2}})\).
Integral is \(\frac{1}{3} \cdot \frac{1}{\sqrt{2}/3} \tan^{-1}(\frac{3x+1}{\sqrt{2}})\).
\(I = \frac{5}{6} \log|1+2x+3x^2| - \frac{11}{3\sqrt{2}} \tan^{-1}\left(\frac{3x+1}{\sqrt{2}}\right) + C\)
Question 19: Integrate the function \(\frac{6x+7}{\sqrt{(x-5)(x-4)}}\)
Denominator is \(\sqrt{x^2-9x+20}\). Derivative is \(2x-9\).
Let \(6x+7 = A(2x-9) + B\).
\(2A = 6 \Rightarrow A = 3\).
\(-9A + B = 7 \Rightarrow -27 + B = 7 \Rightarrow B = 34\).
Let \(6x+7 = A(2x-9) + B\).
\(2A = 6 \Rightarrow A = 3\).
\(-9A + B = 7 \Rightarrow -27 + B = 7 \Rightarrow B = 34\).
\(I = 3 \int \frac{2x-9}{\sqrt{x^2-9x+20}} dx + 34 \int \frac{dx}{\sqrt{x^2-9x+20}}\)
First part: \(3 \cdot 2\sqrt{x^2-9x+20} = 6\sqrt{x^2-9x+20}\).
Second part: Complete square \(\sqrt{(x-9/2)^2 - (1/2)^2}\).
Integral is \(34 \log|x-9/2 + \sqrt{x^2-9x+20}|\).
Second part: Complete square \(\sqrt{(x-9/2)^2 - (1/2)^2}\).
Integral is \(34 \log|x-9/2 + \sqrt{x^2-9x+20}|\).
\(I = 6\sqrt{x^2-9x+20} + 34\log|x-\frac{9}{2} + \sqrt{x^2-9x+20}| + C\)
Question 20: Integrate the function \(\frac{x+2}{\sqrt{4x-x^2}}\)
Derivative of \(4x-x^2\) is \(4-2x\).
Let \(x+2 = A(4-2x) + B\).
\(-2A = 1 \Rightarrow A = -1/2\).
\(4A + B = 2 \Rightarrow -2 + B = 2 \Rightarrow B = 4\).
Let \(x+2 = A(4-2x) + B\).
\(-2A = 1 \Rightarrow A = -1/2\).
\(4A + B = 2 \Rightarrow -2 + B = 2 \Rightarrow B = 4\).
\(I = -\frac{1}{2} \int \frac{4-2x}{\sqrt{4x-x^2}} dx + 4 \int \frac{dx}{\sqrt{4x-x^2}}\)
First part: \(-\frac{1}{2} \cdot 2\sqrt{4x-x^2} = -\sqrt{4x-x^2}\).
Second part: \(4x-x^2 = -(x^2-4x+4-4) = 4-(x-2)^2\).
Integral is \(4 \sin^{-1}(\frac{x-2}{2})\).
Second part: \(4x-x^2 = -(x^2-4x+4-4) = 4-(x-2)^2\).
Integral is \(4 \sin^{-1}(\frac{x-2}{2})\).
\(I = -\sqrt{4x-x^2} + 4\sin^{-1}\left(\frac{x-2}{2}\right) + C\)
Question 21: Integrate the function \(\frac{x+2}{\sqrt{x^2+2x+3}}\)
Derivative of \(x^2+2x+3\) is \(2x+2\).
Rewrite \(x+2 = \frac{1}{2}(2x+2) + 1\).
Rewrite \(x+2 = \frac{1}{2}(2x+2) + 1\).
\(I = \frac{1}{2} \int \frac{2x+2}{\sqrt{x^2+2x+3}} dx + \int \frac{dx}{\sqrt{x^2+2x+3}}\)
First part: \(\sqrt{x^2+2x+3}\).
Second part: Complete square \(\sqrt{(x+1)^2+2}\).
Integral is \(\log|x+1 + \sqrt{x^2+2x+3}|\).
Second part: Complete square \(\sqrt{(x+1)^2+2}\).
Integral is \(\log|x+1 + \sqrt{x^2+2x+3}|\).
\(I = \sqrt{x^2+2x+3} + \log|x+1+\sqrt{x^2+2x+3}| + C\)
Question 22: Integrate the function \(\frac{x+3}{x^2-2x-5}\)
Derivative of \(x^2-2x-5\) is \(2x-2\).
Let \(x+3 = \frac{1}{2}(2x-2) + 4\).
Let \(x+3 = \frac{1}{2}(2x-2) + 4\).
\(I = \frac{1}{2} \int \frac{2x-2}{x^2-2x-5} dx + 4 \int \frac{dx}{x^2-2x-5}\)
First part: \(\frac{1}{2} \log|x^2-2x-5|\).
Second part: Complete square \((x-1)^2 - 6 = (x-1)^2 - (\sqrt{6})^2\).
Using formula \(\frac{1}{2a}\log|\frac{x-a}{x+a}|\):
\(4 \cdot \frac{1}{2\sqrt{6}} \log|\frac{x-1-\sqrt{6}}{x-1+\sqrt{6}}|\).
Second part: Complete square \((x-1)^2 - 6 = (x-1)^2 - (\sqrt{6})^2\).
Using formula \(\frac{1}{2a}\log|\frac{x-a}{x+a}|\):
\(4 \cdot \frac{1}{2\sqrt{6}} \log|\frac{x-1-\sqrt{6}}{x-1+\sqrt{6}}|\).
\(I = \frac{1}{2}\log|x^2-2x-5| + \frac{2}{\sqrt{6}}\log\left|\frac{x-1-\sqrt{6}}{x-1+\sqrt{6}}\right| + C\)
Question 23: Integrate the function \(\frac{5x+3}{\sqrt{x^2+4x+10}}\)
Derivative of \(x^2+4x+10\) is \(2x+4\).
Let \(5x+3 = A(2x+4) + B \Rightarrow 2A=5 \Rightarrow A=5/2\).
\(4A+B=3 \Rightarrow 10+B=3 \Rightarrow B=-7\).
Let \(5x+3 = A(2x+4) + B \Rightarrow 2A=5 \Rightarrow A=5/2\).
\(4A+B=3 \Rightarrow 10+B=3 \Rightarrow B=-7\).
\(I = \frac{5}{2} \int \frac{2x+4}{\sqrt{x^2+4x+10}} dx - 7 \int \frac{dx}{\sqrt{x^2+4x+10}}\)
First part: \(5\sqrt{x^2+4x+10}\).
Second part: Complete square \(\sqrt{(x+2)^2 + 6}\).
Integral is \(-7 \log|x+2 + \sqrt{x^2+4x+10}|\).
Second part: Complete square \(\sqrt{(x+2)^2 + 6}\).
Integral is \(-7 \log|x+2 + \sqrt{x^2+4x+10}|\).
\(I = 5\sqrt{x^2+4x+10} - 7\log|x+2+\sqrt{x^2+4x+10}| + C\)
Question 24: \(\int \frac{dx}{x^2+2x+2}\) equals
\(x^2+2x+2 = (x+1)^2 + 1\).
Integral is \(\int \frac{dx}{(x+1)^2+1} = \tan^{-1}(x+1) + C\).
Integral is \(\int \frac{dx}{(x+1)^2+1} = \tan^{-1}(x+1) + C\).
Correct Answer: B
Question 25: \(\int \frac{dx}{\sqrt{9x-4x^2}}\) equals
Factor 4 from under the root: \(\sqrt{4(\frac{9}{4}x-x^2)} = 2\sqrt{\frac{9}{4}x-x^2}\).
\(I = \frac{1}{2} \int \frac{dx}{\sqrt{- (x^2 - \frac{9}{4}x)}} = \frac{1}{2} \int \frac{dx}{\sqrt{(\frac{9}{8})^2 - (x-\frac{9}{8})^2}}\).
\(I = \frac{1}{2} \int \frac{dx}{\sqrt{- (x^2 - \frac{9}{4}x)}} = \frac{1}{2} \int \frac{dx}{\sqrt{(\frac{9}{8})^2 - (x-\frac{9}{8})^2}}\).
Using \(\sin^{-1}(x/a)\) formula:
\(\frac{1}{2} \sin^{-1}\left( \frac{x-9/8}{9/8} \right) = \frac{1}{2} \sin^{-1}\left( \frac{8x-9}{9} \right)\).
\(\frac{1}{2} \sin^{-1}\left( \frac{x-9/8}{9/8} \right) = \frac{1}{2} \sin^{-1}\left( \frac{8x-9}{9} \right)\).
Correct Answer: B