CBSE Class 12 Mathematics Integrals Worksheet Set 03

Question. Find \( \int \frac{\sqrt{x}}{\sqrt{a^3 - x^3}} \, dx \).
Answer: Let \( I = \int \frac{\sqrt{x}}{\sqrt{a^3 - x^3}} \, dx \)
Put \( x^{3/2} = t \Rightarrow \frac{3}{2} x^{1/2} \, dx = dt \)
\( \therefore I = \frac{2}{3} \int \frac{dt}{\sqrt{(a^{3/2})^2 - t^2}} \)
\( = \frac{2}{3} \left[ \sin^{-1}\left( \frac{t}{a^{3/2}} \right) \right] + C \)
\( = \frac{2}{3} \sin^{-1}\left( \frac{x^{3/2}}{a^{3/2}} \right) + C \)
\( = \frac{2}{3} \sin^{-1}\left( \frac{x}{a} \right)^{3/2} + C \)

Question. Find \( \int (\sqrt{\tan x} + \sqrt{\cot x}) \, dx \).
Answer: Let \( I = \int (\sqrt{\tan x} + \sqrt{\cot x}) \, dx \)
\( = \int \left( \sqrt{\frac{\sin x}{\cos x}} + \sqrt{\frac{\cos x}{\sin x}} \right) dx = \int \frac{\sin x + \cos x}{\sqrt{\sin x \cos x}} \, dx \)
\( = \sqrt{2} \int \frac{\sin x + \cos x}{\sqrt{2\sin x \cos x}} \, dx = \sqrt{2} \int \frac{\sin x + \cos x}{\sqrt{\sin 2x}} \, dx \)
\( = \sqrt{2} \int \frac{\sin x + \cos x}{\sqrt{1 - (1 - \sin 2x)}} \, dx = \sqrt{2} \int \frac{\sin x + \cos x}{\sqrt{1 - (\sin x - \cos x)^2}} \, dx \)
Put \( \sin x - \cos x = t \Rightarrow (\cos x + \sin x) \, dx = dt \)
\( \therefore I = \sqrt{2} \int \frac{dt}{\sqrt{1 - t^2}} = \sqrt{2} \sin^{-1} t + C \)
\( = \sqrt{2} \sin^{-1}(\sin x - \cos x) + C \)

Question. Evaluate : \( \int \frac{x+2}{2x^2 + 6x + 5} \, dx \).
Answer: Let \( I = \int \frac{x+2}{2x^2 + 6x + 5} \, dx \)
We write, \( x + 2 = A \left[ \frac{d}{dx}(2x^2 + 6x + 5) \right] + B \)
\( \Rightarrow x + 2 = A(4x + 6) + B \)
Equating coefficients of \( x \) and constant terms, we get
\( 4A = 1 \Rightarrow A = 1/4 \) and \( 6A + B = 2 \Rightarrow B = 1/2 \)
\( \therefore I = \frac{1}{4} \int \frac{4x+6}{2x^2 + 6x + 5} \, dx + \frac{1}{2} \int \frac{dx}{2x^2 + 6x + 5} \)
\( = \frac{1}{4} \log |2x^2 + 6x + 5| + \frac{1}{4} \int \frac{dx}{x^2 + 3x + \frac{5}{2}} \)
\( = \frac{1}{4} \log |2x^2 + 6x + 5| + \frac{1}{4} \int \frac{dx}{x^2 + 3x + \frac{9}{4} - \frac{9}{4} + \frac{5}{2}} \)
\( = \frac{1}{4} \log |2x^2 + 6x + 5| + \frac{1}{4} \int \frac{dx}{\left(x + \frac{3}{2}\right)^2 + \left(\frac{1}{2}\right)^2} \)
\( = \frac{1}{4} \log |2x^2 + 6x + 5| + \frac{1}{4} \cdot \frac{1}{\left(\frac{1}{2}\right)} \tan^{-1}\left( \frac{x + \frac{3}{2}}{\frac{1}{2}} \right) + C \)
\( = \frac{1}{4} \log |2x^2 + 6x + 5| + \frac{1}{2} \tan^{-1}(2x + 3) + C \)

Question. Find \( \int \frac{x+3}{\sqrt{5 - 4x - 2x^2}} \, dx \).
Answer: Let \( I = \int \frac{x+3}{\sqrt{5-4x-2x^2}} \, dx \)
Let \( x + 3 = A \frac{d}{dx}(5-4x-2x^2) + B = A(-4-4x) + B \)
On comparing the like coefficients, we get
\( x + 3 = -\frac{1}{4}(-4-4x) + 2 \)
\( \Rightarrow I = \int \frac{-\frac{1}{4}(-4-4x) + 2}{\sqrt{5-4x-2x^2}} \, dx \)
\( = -\frac{1}{4} \int \frac{-4-4x}{\sqrt{5-4x-2x^2}} \, dx + 2 \int \frac{1}{\sqrt{5-4x-2x^2}} \, dx \)
\( \Rightarrow I = -\frac{1}{4}I_1 + 2I_2 \) ...(1)
where \( I_1 = \int \frac{-4-4x}{\sqrt{5-4x-2x^2}} \, dx \)
Put \( 5 - 4x - 2x^2 = t \Rightarrow (-4-4x)dx = dt \)
\( \therefore I_1 = \int \frac{dt}{\sqrt{t}} = \int t^{-1/2} dt = 2\sqrt{t} + C_1 = 2\sqrt{5-4x-2x^2} + C_1 \) ...(2)
and \( I_2 = \int \frac{dx}{\sqrt{5-4x-2x^2}} = \frac{1}{\sqrt{2}} \int \frac{dx}{\sqrt{\frac{5}{2} - 2x - x^2}} \)
\( = \frac{1}{\sqrt{2}} \int \frac{dx}{\sqrt{\frac{7}{2} - (x^2+2x+1)}} = \frac{1}{\sqrt{2}} \int \frac{dx}{\sqrt{\left(\frac{\sqrt{7}}{\sqrt{2}}\right)^2 - (x+1)^2}} \)
\( = \frac{1}{\sqrt{2}} \sin^{-1} \left( \frac{x+1}{\sqrt{7/2}} \right) + C_2 = \frac{1}{\sqrt{2}} \sin^{-1}\left[ \sqrt{\frac{2}{7}}(x+1) \right] + C_2 \) ...(3)
From (1), (2) and (3), we get
\( I = -\frac{1}{4} \cdot 2\sqrt{5-4x-2x^2} + 2 \cdot \frac{1}{\sqrt{2}} \sin^{-1}\left[ \sqrt{\frac{2}{7}}(x+1) \right] + C \)
where \( C = C_1 + C_2 \)
\( \Rightarrow I = -\frac{1}{2} \sqrt{5-4x-2x^2} + \sqrt{2} \sin^{-1}\left[ \sqrt{\frac{2}{7}}(x+1) \right] + C \)

Question. Evaluate : \( \int \frac{x+2}{\sqrt{x^2+5x+6}} \, dx \).
Answer: Let \( I = \int \frac{x+2}{\sqrt{x^2+5x+6}} \, dx = \int \frac{\frac{1}{2}(2x+5) - \frac{1}{2}}{\sqrt{x^2+5x+6}} \, dx \)
\( = \frac{1}{2} \int (x^2+5x+6)^{-1/2}(2x+5)\,dx - \frac{1}{2} \int \frac{dx}{\sqrt{x^2+5x+6}} \)
Put \( x^2 + 5x + 6 = t \Rightarrow (2x + 5) \, dx = dt \)
\( \Rightarrow I = \frac{1}{2} \int t^{-1/2} \, dt - \frac{1}{2} \int \frac{dx}{\sqrt{\left(x + \frac{5}{2}\right)^2 - \left(\frac{1}{2}\right)^2}} \)
\( = \frac{1}{2} \frac{t^{1/2}}{1/2} - \frac{1}{2} \log \left| \left(x + \frac{5}{2}\right) + \sqrt{\left(x + \frac{5}{2}\right)^2 - \left(\frac{1}{2}\right)^2} \right| + C \)
\( = \sqrt{x^2+5x+6} - \frac{1}{2} \log \left| x + \frac{5}{2} + \sqrt{x^2+5x+6} \right| + C \)

Question. Evaluate : \( \int \frac{5x-2}{1+2x+3x^2} \, dx \).
Answer: Let \( I = \int \frac{5x-2}{1+2x+3x^2} \, dx \)
Let \( 5x-2 = A \frac{d}{dx}(1+2x+3x^2) + B = A(2+6x) + B \)
On comparing the like coefficients, we get
\( 5 = 6A \Rightarrow A = \frac{5}{6} \) and \( 2A + B = -2 \Rightarrow 2\left(\frac{5}{6}\right) + B = -2 \Rightarrow B = -2 - \frac{5}{3} = -\frac{11}{3} \)
So, \( 5x-2 = \frac{5}{6}(2+6x) - \frac{11}{3} \)
\( \therefore I = \int \frac{\frac{5}{6}(2+6x) - \frac{11}{3}}{1+2x+3x^2} \, dx \)
\( = \frac{5}{6} \int \frac{2+6x}{1+2x+3x^2} \, dx - \frac{11}{3} \int \frac{dx}{1+2x+3x^2} \)
\( = \frac{5}{6} \log |1+2x+3x^2| - \frac{11}{3 \cdot 3} \int \frac{dx}{x^2 + \frac{2}{3}x + \frac{1}{3}} \)
\( = \frac{5}{6} \log |1+2x+3x^2| - \frac{11}{9} \int \frac{dx}{\left(x + \frac{1}{3}\right)^2 + \left(\frac{\sqrt{2}}{3}\right)^2} \)
\( = \frac{5}{6} \log |1+2x+3x^2| - \frac{11}{9} \cdot \frac{1}{\left(\frac{\sqrt{2}}{3}\right)} \tan^{-1}\left( \frac{x + \frac{1}{3}}{\frac{\sqrt{2}}{3}} \right) + C \)
\( = \frac{5}{6} \log |1+2x+3x^2| - \frac{11}{3\sqrt{2}} \tan^{-1}\left( \frac{3x+1}{\sqrt{2}} \right) + C \)

Question. Evaluate : \( \int \frac{x+2}{\sqrt{x^2+2x+3}} \, dx \).
Answer: Let \( I = \int \frac{x+2}{\sqrt{x^2+2x+3}} \, dx = \int \frac{(x+1)+1}{\sqrt{x^2+2x+3}} \, dx \)
Put \( x^2 + 2x + 3 = t \Rightarrow (2x+2)dx = dt \Rightarrow (x+1)dx = \frac{1}{2}dt \)
\( \therefore I = \frac{1}{2} \int t^{-1/2} \, dt + \int \frac{1}{\sqrt{(x+1)^2 + (\sqrt{2})^2}} \, dx \)
\( = \frac{1}{2} \frac{t^{1/2}}{1/2} + \log\left| (x+1) + \sqrt{x^2+2x+3} \right| + C \)
\( = \sqrt{x^2+2x+3} + \log\left| (x+1) + \sqrt{x^2+2x+3} \right| + C \)

Question. Evaluate : \( \int \frac{5x+3}{\sqrt{x^2+4x+10}} \, dx \).
Answer: Let \( I = \int \frac{5x+3}{\sqrt{x^2+4x+10}} \, dx = \int \frac{\frac{5}{2}(2x+4) - 7}{\sqrt{x^2+4x+10}} \, dx \)
\( = \frac{5}{2} \int \frac{2x+4}{\sqrt{x^2+4x+10}} \, dx - 7 \int \frac{dx}{\sqrt{x^2+4x+10}} \)
\( = I_1 + I_2 \) (say) ...(1)
where \( I_1 = \frac{5}{2} \int \frac{2x+4}{\sqrt{x^2+4x+10}} \, dx \)
Put \( x^2+4x+10 = t \Rightarrow (2x+4)dx = dt \)
\( \therefore I_1 = \frac{5}{2} \int t^{-1/2} \, dt = \frac{5}{2} \frac{t^{1/2}}{1/2} = 5\sqrt{t} = 5\sqrt{x^2+4x+10} + C_1 \) ...(2)
and \( I_2 = -7 \int \frac{dx}{\sqrt{x^2+4x+10}} = -7 \int \frac{dx}{\sqrt{(x+2)^2 + (\sqrt{6})^2}} \)
\( = -7 \log \left| x+2 + \sqrt{x^2+4x+10} \right| + C_2 \) ...(3)
From (1), (2) and (3), we get
\( I = 5\sqrt{x^2+4x+10} - 7 \log \left| x+2 + \sqrt{x^2+4x+10} \right| + C \), where \( C = C_1 + C_2 \)

Question. Evaluate : \( \int \frac{\sin x - \cos x}{\sqrt{\sin 2x}} \, dx \).
Answer: Let \( I = \int \frac{\sin x - \cos x}{\sqrt{\sin 2x}} \, dx = \int \frac{\sin x - \cos x}{\sqrt{1 + \sin 2x - 1}} \, dx \)
\( = \int \frac{\sin x - \cos x}{\sqrt{\sin^2 x + \cos^2 x + 2\sin x \cos x - 1}} \, dx = \int \frac{\sin x - \cos x}{\sqrt{(\sin x + \cos x)^2 - 1}} \, dx \)
Put \( \sin x + \cos x = t \Rightarrow (\cos x - \sin x)dx = dt \Rightarrow (\sin x - \cos x)dx = -dt \)
\( \dots I = \int \frac{-dt}{\sqrt{t^2 - 1}} = -\log\left| t + \sqrt{t^2 - 1} \right| + C \)
\( = -\log\left| \sin x + \cos x + \sqrt{\sin 2x} \right| + C \)

Question. Evaluate : \( \int \frac{x^2+1}{x^4+1} \, dx \).
Answer: Let \( I = \int \frac{x^2+1}{x^4+1} \, dx = \int \frac{1 + \frac{1}{x^2}}{x^2 + \frac{1}{x^2}} \, dx = \int \frac{1 + \frac{1}{x^2}}{\left(x - \frac{1}{x}\right)^2 + 2} \, dx \)
Put \( x - \frac{1}{x} = t \Rightarrow \left(1 + \frac{1}{x^2}\right)dx = dt \)
\( \therefore I = \int \frac{dt}{t^2 + (\sqrt{2})^2} = \frac{1}{\sqrt{2}} \tan^{-1}\left( \frac{t}{\sqrt{2}} \right) + C \)
\( = \frac{1}{\sqrt{2}} \tan^{-1}\left( \frac{x - \frac{1}{x}}{\sqrt{2}} \right) + C = \frac{1}{\sqrt{2}} \tan^{-1}\left( \frac{x^2 - 1}{\sqrt{2}x} \right) + C \)

Question. Evaluate : \( \int \frac{x^2+4}{x^4+16} \, dx \).
Answer: Let \( I = \int \frac{x^2+4}{x^4+16} \, dx \)
\( \Rightarrow I = \int \frac{1 + \frac{4}{x^2}}{x^2 + \frac{16}{x^2}} \, dx = \int \frac{1 + \frac{4}{x^2}}{\left(x - \frac{4}{x}\right)^2 + 8} \, dx \)
Put \( x - \frac{4}{x} = t \Rightarrow \left(1 + \frac{4}{x^2}\right)dx = dt \)
\( \therefore I = \int \frac{dt}{t^2 + 8} = \int \frac{dt}{t^2 + (2\sqrt{2})^2} \)
\( = \frac{1}{2\sqrt{2}} \tan^{-1}\left( \frac{t}{2\sqrt{2}} \right) + C = \frac{1}{2\sqrt{2}} \tan^{-1}\left( \frac{x - \frac{4}{x}}{2\sqrt{2}} \right) + C \)
\( = \frac{1}{2\sqrt{2}} \tan^{-1}\left( \frac{x^2 - 4}{2\sqrt{2}x} \right) + C \)

Question. Evaluate the following : \( \int \frac{x+2}{\sqrt{(x-2)(x-3)}} \, dx \).
Answer: Let \( I = \int \frac{x+2}{\sqrt{(x-2)(x-3)}} \, dx = \int \frac{x+2}{\sqrt{x^2-5x+6}} \, dx \)
We write, \( x + 2 = A \left[ \frac{d}{dx}(x^2 - 5x + 6) \right] + B = A(2x-5) + B \)
Equating coefficients of \( x \) and constant terms, we get
\( 2A = 1 \Rightarrow A = \frac{1}{2} \) and \( -5A + B = 2 \Rightarrow B = \frac{9}{2} \)
\( \therefore I = \frac{1}{2} \int \frac{2x-5}{\sqrt{x^2-5x+6}} \, dx + \frac{9}{2} \int \frac{dx}{\sqrt{x^2-5x+6}} \)
\( = \frac{1}{2} \cdot 2\sqrt{x^2-5x+6} + \frac{9}{2} \int \frac{dx}{\sqrt{x^2 - 5x + \frac{25}{4} - \frac{25}{4} + 6}} \)
\( = \sqrt{x^2-5x+6} + \frac{9}{2} \int \frac{dx}{\sqrt{\left(x - \frac{5}{2}\right)^2 - \left(\frac{1}{2}\right)^2}} \)
\( = \sqrt{x^2-5x+6} + \frac{9}{2} \log \left| \left(x - \frac{5}{2}\right) + \sqrt{x^2-5x+6} \right| + C \)

Question. Evaluate : \( \int \frac{dx}{\sqrt{5 - 4x - 2x^2}} \).
Answer: Let \( I = \int \frac{dx}{\sqrt{5-4x-2x^2}} = \frac{1}{\sqrt{2}} \int \frac{dx}{\sqrt{\frac{5}{2} - 2x - x^2}} \)
\( = \frac{1}{\sqrt{2}} \int \frac{dx}{\sqrt{\frac{7}{2} - (x^2 + 2x + 1)}} = \frac{1}{\sqrt{2}} \int \frac{dx}{\sqrt{\left(\frac{\sqrt{7}}{\sqrt{2}}\right)^2 - (x+1)^2}} \)
\( = \frac{1}{\sqrt{2}} \sin^{-1} \left( \frac{x+1}{\sqrt{7/2}} \right) + C = \frac{1}{\sqrt{2}} \sin^{-1}\left[ \sqrt{\frac{2}{7}}(x+1) \right] + C \)

Question. Evaluate : \( \int \frac{2x+5}{\sqrt{7-6x-x^2}} \, dx \).
Answer: Let \( I = \int \frac{2x+5}{\sqrt{7-6x-x^2}} \, dx \)
We write, \( 2x+5 = A \left[ \frac{d}{dx}(7-6x-x^2) \right] + B = A(-6-2x) + B \)
Equating coefficients of \( x \) and constant terms, we get
\( 2 = -2A \Rightarrow A = -1 \) and \( -6A + B = 5 \Rightarrow B = -1 \)
\( \dots I = \int \frac{-(-6-2x) - 1}{\sqrt{7-6x-x^2}} \, dx \)
\( = -\int \frac{-6-2x}{\sqrt{7-6x-x^2}} \, dx - \int \frac{dx}{\sqrt{7-6x-x^2}} \)
\( = -2\sqrt{7-6x-x^2} - \int \frac{dx}{\sqrt{7 + 9 - (9 + 6x + x^2)}} \)
\( = -2\sqrt{7-6x-x^2} - \int \frac{dx}{\sqrt{16 - (x+3)^2}} \)
\( = -2\sqrt{7-6x-x^2} - \int \frac{dx}{\sqrt{4^2 - (x+3)^2}} \)
\( = -2\sqrt{7-6x-x^2} - \sin^{-1}\left( \frac{x+3}{4} \right) + C \)

Question. Evaluate : \( \int \frac{\sin x + \cos x}{\sqrt{\sin 2x}} \, dx \).
Answer: Let \( I = \int \frac{\sin x + \cos x}{\sqrt{\sin 2x}} \, dx \)
\( = \int \frac{\sin x + \cos x}{\sqrt{1 - (1 - \sin 2x)}} \, dx = \int \frac{\sin x + \cos x}{\sqrt{1 - (\sin x - \cos x)^2}} \, dx \)
Put \( \sin x - \cos x = t \Rightarrow (\cos x + \sin x) \, dx = dt \)
\( \therefore I = \int \frac{dt}{\sqrt{1-t^2}} = \sin^{-1} t + C = \sin^{-1}(\sin x - \cos x) + C \)

Question. Evaluate : \( \int \frac{x+3}{x^2-2x-5} \, dx \).
Answer: Let \( I = \int \frac{x+3}{x^2-2x-5} \, dx \)
We write, \( x + 3 = A \frac{d}{dx}(x^2 - 2x - 5) + B = A(2x-2) + B \)
Equating coefficients of \( x \) and constant terms, we get
\( 2A = 1 \Rightarrow A = 1/2 \) and \( -2A + B = 3 \Rightarrow B = 4 \)
\( \therefore I = \frac{1}{2} \int \frac{2x-2}{x^2-2x-5} \, dx + 4 \int \frac{dx}{x^2-2x-5} \)
\( = \frac{1}{2} \log |x^2-2x-5| + 4 \int \frac{dx}{x^2-2x+1-6} \)
\( = \frac{1}{2} \log |x^2-2x-5| + 4 \int \frac{dx}{(x-1)^2 - (\sqrt{6})^2} \)
\( = \frac{1}{2} \log |x^2-2x-5| + \frac{4}{2\sqrt{6}} \log \left| \frac{x-1-\sqrt{6}}{x-1+\sqrt{6}} \right| + C \)
\( = \frac{1}{2} \log |x^2-2x-5| + \sqrt{\frac{2}{3}} \log \left| \frac{x-1-\sqrt{6}}{x-1+\sqrt{6}} \right| + C \)

Question. Evaluate : \( \int \frac{1-x^2}{1+x^4} \, dx \).
Answer: Let \( I = \int \frac{1-x^2}{1+x^4} \, dx = \int \frac{\frac{1}{x^2} - 1}{x^2 + \frac{1}{x^2}} \, dx = -\int \frac{1 - \frac{1}{x^2}}{\left(x + \frac{1}{x}\right)^2 - 2} \, dx \)
Put \( x + \frac{1}{x} = t \Rightarrow \left(1 - \frac{1}{x^2}\right)dx = dt \)
\( \therefore I = -\int \frac{dt}{t^2 - (\sqrt{2})^2} = -\frac{1}{2\sqrt{2}} \log \left| \frac{t - \sqrt{2}}{t + \sqrt{2}} \right| + C \)
\( = -\frac{1}{2\sqrt{2}} \log \left| \frac{x + \frac{1}{x} - \sqrt{2}}{x + \frac{1}{x} + \sqrt{2}} \right| + C = -\frac{1}{2\sqrt{2}} \log \left| \frac{x^2 - \sqrt{2}x + 1}{x^2 + \sqrt{2}x + 1} \right| + C \)

Question. Evaluate : \( \int \frac{x}{x^2+x+1} \, dx \).
Answer: Let \( I = \int \frac{x}{x^2+x+1} \, dx \)
We write, \( x = A \left[ \frac{d}{dx}(x^2 + x + 1) \right] + B = A[2x+1] + B \)
Equating coefficients of \( x \) and constant terms, we get
\( 2A = 1 \Rightarrow A = 1/2 \) and \( A + B = 0 \Rightarrow B = -1/2 \)
\( \therefore I = \frac{1}{2} \int \frac{2x+1}{x^2+x+1} \, dx - \frac{1}{2} \int \frac{dx}{x^2+x+1} \)
\( = \frac{1}{2} \log|x^2+x+1| - \frac{1}{2} \int \frac{dx}{x^2 + x + \frac{1}{4} + \frac{3}{4}} \)
\( = \frac{1}{2} \log|x^2+x+1| - \frac{1}{2} \int \frac{dx}{\left(x + \frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2} \)
\( = \frac{1}{2} \log|x^2+x+1| - \frac{1}{2} \cdot \frac{1}{\left(\frac{\sqrt{3}}{2}\right)} \tan^{-1}\left( \frac{x+\frac{1}{2}}{\frac{\sqrt{3}}{2}} \right) + C \)
\( = \frac{1}{2} \log|x^2+x+1| - \frac{1}{\sqrt{3}} \tan^{-1}\left( \frac{2x+1}{\sqrt{3}} \right) + C \)

Question. Evaluate : \( \int \frac{1}{\cos^4 x + \sin^4 x} \, dx \).
Answer: Let \( I = \int \frac{1}{\cos^4 x + \sin^4 x} \, dx = \int \frac{\sec^4 x}{1 + \tan^4 x} \, dx \)
\( = \int \frac{(\tan^2 x + 1)\sec^2 x}{1 + \tan^4 x} \, dx \)
Put \( \tan x = t \Rightarrow \sec^2 x \, dx = dt \)
\( \therefore I = \int \frac{t^2+1}{t^4+1} \, dt = \int \frac{1 + \frac{1}{t^2}}{t^2 + \frac{1}{t^2}} \, dt = \int \frac{1 + \frac{1}{t^2}}{\left(t - \frac{1}{t}\right)^2 + 2} \, dt \)
Put \( t - \frac{1}{t} = y \Rightarrow \left(1 + \frac{1}{t^2}\right) \, dt = dy \)
\( \therefore I = \int \frac{dy}{y^2 + (\sqrt{2})^2} = \frac{1}{\sqrt{2}} \tan^{-1}\left(\frac{y}{\sqrt{2}}\right) + C \)
\( = \frac{1}{\sqrt{2}} \tan^{-1}\left( \frac{\tan x - \cot x}{\sqrt{2}} \right) + C \)

Question. Evaluate : \( \int \frac{1}{\sin^4 x + \sin^2 x \cos^2 x + \cos^4 x} \, dx \).
Answer: Let \( I = \int \frac{1}{\sin^4 x + \sin^2 x \cos^2 x + \cos^4 x} \, dx = \int \frac{\sec^4 x}{\tan^4 x + \tan^2 x + 1} \, dx \)
\( = \int \frac{(\tan^2 x + 1)\sec^2 x}{\tan^4 x + \tan^2 x + 1} \, dx \)
Put \( \tan x = t \Rightarrow \sec^2 x \, dx = dt \)
\( I = \int \frac{t^2+1}{t^4 + t^2 + 1} \, dt = \int \frac{1 + \frac{1}{t^2}}{t^2 + 1 + \frac{1}{t^2}} \, dt = \int \frac{1 + \frac{1}{t^2}}{\left(t - \frac{1}{t}\right)^2 + 3} \, dt \)
Put \( t - \frac{1}{t} = y \Rightarrow \left(1 + \frac{1}{t^2}\right) \, dt = dy \)
Thus, \( I = \int \frac{dy}{y^2 + (\sqrt{3})^2} = \frac{1}{\sqrt{3}} \tan^{-1}\left(\frac{y}{\sqrt{3}}\right) + C \)
\( = \frac{1}{\sqrt{3}} \tan^{-1}\left( \frac{\tan x - \cot x}{\sqrt{3}} \right) + C \)

Question. Evaluate : \( \int \frac{6x+7}{\sqrt{(x-5)(x-4)}} \, dx \).
Answer: Let \( I = \int \frac{6x+7}{\sqrt{(x-5)(x-4)}} \, dx = \int \frac{6x+7}{\sqrt{x^2-9x+20}} \, dx \)
Let \( 6x + 7 = A \left[ \frac{d}{dx}(x^2 - 9x + 20) \right] + B \)
\( \therefore 6x + 7 = A[2x - 9] + B \)
Equating the coefficients of like terms from both sides, we get
\( 2A = 6 \Rightarrow A = 3 \) and \( -9A + B = 7 \Rightarrow B = 7 + 27 = 34 \)
\( \therefore I = \int \frac{3(2x-9)}{\sqrt{x^2-9x+20}} \, dx + \int \frac{34}{\sqrt{x^2-9x+20}} \, dx \)
Put \( x^2 - 9x + 20 = t \) in the first integral
\( \therefore I = \int 3t^{-1/2} \, dt + 34 \int \frac{dx}{\sqrt{\left(x - \frac{9}{2}\right)^2 - \frac{81}{4} + 20}} \)
\( = 3(2\sqrt{t}) + 34 \int \frac{dx}{\sqrt{\left(x - \frac{9}{2}\right)^2 - \frac{1}{4}}} \)
\( = 6\sqrt{x^2-9x+20} + 34 \log \left| \left(x - \frac{9}{2}\right) + \sqrt{\left(x - \frac{9}{2}\right)^2 - \frac{1}{4}} \right| + C \)
\( = 6\sqrt{x^2-9x+20} + 34 \log \left| \left(x - \frac{9}{2}\right) + \sqrt{x^2-9x+20} \right| + C \)

Question. Find : \( \int \frac{x^2}{x^4+x^2-2} \, dx \).
Answer: Let \( I = \int \frac{x^2}{x^4+x^2-2} \, dx = \int \frac{x^2}{(x^2-1)(x^2+2)} \, dx \)
Let \( x^2 = z \)
\( \therefore \frac{x^2}{(x^2-1)(x^2+2)} = \frac{z}{(z-1)(z+2)} \)
Using partial fractions, we have
\( \frac{z}{(z-1)(z+2)} = \frac{A}{z-1} + \frac{B}{z+2} \)
\( \Rightarrow z = A(z+2) + B(z-1) \)
When \( z = 1 \), we get \( A = \frac{1}{3} \)
and when \( z = -2 \), we get \( B = \frac{2}{3} \)
\( \therefore I = \int \left[ \frac{1/3}{x^2-1} + \frac{2/3}{x^2+2} \right] dx \)
\( = \frac{1}{3} \int \frac{dx}{x^2-1} + \frac{2}{3} \int \frac{dx}{x^2+(\sqrt{2})^2} \)
\( = \frac{1}{3} \cdot \frac{1}{2} \log\left| \frac{x-1}{x+1} \right| + \frac{2}{3} \cdot \frac{1}{\sqrt{2}} \tan^{-1}\left( \frac{x}{\sqrt{2}} \right) + C \)
\( = \frac{1}{6} \log\left| \frac{x-1}{x+1} \right| + \frac{\sqrt{2}}{3} \tan^{-1}\left( \frac{x}{\sqrt{2}} \right) + C \)

Question. Find : \( \int \frac{(x^2+1)(x^2+4)}{(x^2+3)(x^2-5)} \, dx \).
Answer: Let \( I = \int \frac{(x^2+1)(x^2+4)}{(x^2+3)(x^2-5)} \, dx \)
Let \( x^2 = t \)
\( \dots \frac{(x^2+1)(x^2+4)}{(x^2+3)(x^2-5)} = \frac{(t+1)(t+4)}{(t+3)(t-5)} = \frac{t^2+5t+4}{t^2-2t-15} = 1 + \frac{7t+19}{(t+3)(t-5)} \)
Let \( \frac{7t+19}{(t+3)(t-5)} = \frac{A}{t+3} + \frac{B}{t-5} \)
\( \Rightarrow 7t + 19 = A(t-5) + B(t+3) \)
Putting \( t = 5 \), we get \( B = \frac{27}{4} \)
Putting \( t = -3 \), we get \( A = \frac{1}{4} \)
\( \therefore \frac{t^2+5t+4}{(t+3)(t-5)} = 1 + \frac{1}{4(t+3)} + \frac{27}{4(t-5)} \)
\( \Rightarrow I = \int \left[ 1 + \frac{1}{4(x^2+3)} + \frac{27}{4(x^2-5)} \right] dx \)
\( = x + \frac{1}{4\sqrt{3}} \tan^{-1}\left(\frac{x}{\sqrt{3}}\right) + \frac{27}{4 \cdot 2\sqrt{5}} \log\left| \frac{x-\sqrt{5}}{x+\sqrt{5}} \right| + C \)
\( = x + \frac{1}{4\sqrt{3}} \tan^{-1}\left(\frac{x}{\sqrt{3}}\right) + \frac{27}{8\sqrt{5}} \log\left| \frac{x-\sqrt{5}}{x+\sqrt{5}} \right| + C \)

Question. Find : \( \int \frac{dx}{\sin x + \sin 2x} \).
Answer: Let \( I = \int \frac{dx}{\sin x + \sin 2x} = \int \frac{dx}{\sin x + 2\sin x \cos x} = \int \frac{dx}{\sin x(1+2\cos x)} \)
\( = \int \frac{\sin x \, dx}{\sin^2 x(1+2\cos x)} = \int \frac{\sin x \, dx}{(1-\cos^2 x)(1+2\cos x)} \)
Let \( u = \cos x \Rightarrow du = -\sin x \, dx \)
Also, \( \sin^2 x = 1 - \cos^2 x = 1 - u^2 \)
\( \therefore I = \int \frac{-du}{(1-u^2)(1+2u)} = \int \frac{-du}{(1+u)(1-u)(1+2u)} \)
Using partial fractions, we have
\( \frac{-1}{(1+u)(1-u)(1+2u)} = \frac{A}{1+u} + \frac{B}{1-u} + \frac{C}{1+2u} \)
\( \Rightarrow -1 = A(1-u)(1+2u) + B(1+u)(1+2u) + C(1+u)(1-u) \)
Put \( u = 1 \), we get \( B = -1/6 \)
Put \( u = -1 \), we get \( A = 1/2 \)
Put \( u = -1/2 \), we get \( C = -4/3 \)
So,
\( \frac{-1}{(1+u)(1-u)(1+2u)} = \frac{1}{2(1+u)} - \frac{1}{6(1-u)} - \frac{4}{3(1+2u)} \)
\( \Rightarrow I = \int \left[ \frac{1}{2(1+u)} - \frac{1}{6(1-u)} - \frac{4}{3(1+2u)} \right] du \)
\( = \frac{1}{2} \log(1+u) + \frac{1}{6} \log(1-u) - \frac{4}{3 \cdot 2} \log(1+2u) + C \)
\( = \frac{1}{2} \log(1+\cos x) + \frac{1}{6} \log(1-\cos x) - \frac{2}{3} \log(1+2\cos x) + C \)

Question. Evaluate : \( \int \frac{x^2+x+1}{(x^2+1)(x+2)} \, dx \).
Answer: Let \( I = \int \frac{x^2+x+1}{(x^2+1)(x+2)} \, dx \) ...(1)
Let \( \frac{x^2+x+1}{(x^2+1)(x+2)} = \frac{Ax+B}{x^2+1} + \frac{C}{x+2} \) ...(2)
\( \Rightarrow x^2 + x + 1 = (Ax + B)(x + 2) + C(x^2 + 1) \)
Put \( x = 0, 1, -2 \) in it to get
\( 1 = 2B + C \); \( 3 = 3(A + B) + 2C \) and \( 3 = 5C \)
\( \Rightarrow C = \frac{3}{5}, B = \frac{1}{5} \text{ and } A = \frac{2}{5} \)
Hence, from (2)
\( \frac{x^2+x+1}{(x^2+1)(x+2)} = \frac{\frac{2}{5}x + \frac{1}{5}}{x^2+1} + \frac{3/5}{x+2} = \frac{1}{5} \frac{2x+1}{x^2+1} + \frac{3}{5} \frac{1}{x+2} \)
\( \therefore I = \frac{1}{5} \int \frac{2x+1}{x^2+1} \, dx + \frac{3}{5} \int \frac{dx}{x+2} \)
\( = \frac{1}{5} \int \frac{2x}{x^2+1} \, dx + \frac{1}{5} \int \frac{dx}{x^2+1} + \frac{3}{5} \int \frac{dx}{x+2} \)
\( = \frac{1}{5} \log|x^2+1| + \frac{1}{5} \tan^{-1}x + \frac{3}{5} \log|x+2| + C_1 \)

Question. Evaluate : \( \int \frac{x^2}{(x^2+4)(x^2+9)} \, dx \).
Answer: Let \( I = \int \frac{x^2}{(x^2+4)(x^2+9)} \, dx \)
Put \( x^2 = y \). Then \( \frac{x^2}{(x^2+4)(x^2+9)} = \frac{y}{(y+4)(y+9)} \)
Let \( \frac{y}{(y+4)(y+9)} = \frac{A}{y+4} + \frac{B}{y+9} \) ...(1)
\( \Rightarrow y = A(y+9) + B(y+4) \) ...(2)
Putting \( y = -4 \) and \( y = -9 \) successively in (2), we get
\( A = -\frac{4}{5} \text{ and } B = \frac{9}{5} \)
Substituting the values of \( A \) and \( B \) in (1), we get
\( \frac{y}{(y+4)(y+9)} = \frac{-4/5}{y+4} + \frac{9/5}{y+9} \)
\( \Rightarrow \frac{x^2}{(x^2+4)(x^2+9)} = -\frac{4}{5(x^2+4)} + \frac{9}{5(x^2+9)} \)
\( \therefore I = \int \frac{x^2}{(x^2+4)(x^2+9)} \, dx = -\frac{4}{5} \int \frac{1}{x^2+4} \, dx + \frac{9}{5} \int \frac{1}{x^2+9} \, dx \)
\( = -\frac{4}{5} \cdot \frac{1}{2} \tan^{-1}\left(\frac{x}{2}\right) + \frac{9}{5} \cdot \frac{1}{3} \tan^{-1}\left(\frac{x}{3}\right) + C \)
\( = -\frac{2}{5} \tan^{-1}\left(\frac{x}{2}\right) + \frac{3}{5} \tan^{-1}\left(\frac{x}{3}\right) + C \)

Question. Find : \( \int \frac{x}{(x-1)^2(x+2)} \, dx \).
Answer: Let \( I = \int \frac{x}{(x-1)^2(x+2)} \, dx \) ...(1)
Let \( \frac{x}{(x-1)^2(x+2)} = \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{C}{x+2} \) ...(2)
\( \Rightarrow x = A(x - 1)(x + 2) + B(x + 2) + C(x - 1)^2 \) ...(3)
Comparing coeffs. of \( x^2 \), \( x \) and constants in (3), we get
\( 0 = A + C \); \( 1 = A + B - 2C \); \( 0 = -2A + 2B + C \)
Solving these, we get
\( A = \frac{2}{9}, B = \frac{1}{3}, C = -\frac{2}{9} \)
\( \therefore \frac{x}{(x-1)^2(x+2)} = \frac{2}{9} \frac{1}{x-1} + \frac{1}{3} \frac{1}{(x-1)^2} - \frac{2}{9} \frac{1}{x+2} \)
\( \dots I = \frac{2}{9} \int \frac{1}{x-1} \, dx + \frac{1}{3} \int \frac{1}{(x-1)^2} \, dx - \frac{2}{9} \int \frac{1}{x+2} \, dx + C_1 \)
\( = \frac{2}{9} \log|x-1| - \frac{1}{3} \frac{1}{x-1} - \frac{2}{9} \log|x+2| + C_1 \)
\( = \frac{2}{9} \log\left| \frac{x-1}{x+2} \right| - \frac{1}{3(x-1)} + C_1 \)

Question. Find \( \int \frac{x}{(x^2+1)(x-1)} \, dx \).
Answer: Let \( I = \int \frac{x}{(x^2+1)(x-1)} \, dx \)
Let \( \frac{x}{(x^2+1)(x-1)} = \frac{Ax+B}{x^2+1} + \frac{C}{x-1} \) ...(1)
\( \Rightarrow x = (Ax + B)(x - 1) + C(x^2 + 1) \) ...(2)
Comparing coefficients of \( x^2 \), \( x \) and constant terms, we get
\( 0 = A + C \); \( 1 = B - A \); \( 0 = -B + C \)
Solving these, we get
\( A = -\frac{1}{2}, C = \frac{1}{2}, B = \frac{1}{2} \)
\( \therefore \text{From (1), we get} \)
\( \frac{x}{(x^2+1)(x-1)} = \frac{-\frac{1}{2}(x-1)}{x^2+1} + \frac{1/2}{x-1} = -\frac{1}{2} \frac{x}{x^2+1} + \frac{1}{2} \frac{1}{x^2+1} + \frac{1}{2} \frac{1}{x-1} \)
\( \therefore I = -\frac{1}{4} \int \frac{2x}{x^2+1} \, dx + \frac{1}{2} \int \frac{dx}{x^2+1} + \frac{1}{2} \int \frac{dx}{x-1} + C \)
\( = -\frac{1}{4} \log|x^2+1| + \frac{1}{2} \tan^{-1}x + \frac{1}{2} \log|x-1| + C \)

Question. Evaluate : \( \int \frac{x^2}{(x^2+1)(x^2+4)} \, dx \).
Answer: Let \( I = \int \frac{x^2}{(x^2+1)(x^2+4)} \, dx \)
Put \( x^2 = y \). Then \( \frac{x^2}{(x^2+1)(x^2+4)} = \frac{y}{(y+1)(y+4)} \)
Let \( \frac{y}{(y+1)(y+4)} = \frac{A}{y+1} + \frac{B}{y+4} \)
\( \Rightarrow y = A(y+4) + B(y+1) \)
Putting \( y = -1 \), we get \( A = -\frac{1}{3} \)
Putting \( y = -4 \), we get \( B = \frac{4}{3} \)
\( \therefore \frac{x^2}{(x^2+1)(x^2+4)} = -\frac{1}{3(x^2+1)} + \frac{4}{3(x^2+4)} \)
\( \therefore I = -\frac{1}{3} \int \frac{dx}{x^2+1} + \frac{4}{3} \int \frac{dx}{x^2+4} \)
\( = -\frac{1}{3} \tan^{-1}x + \frac{4}{3} \cdot \frac{1}{2} \tan^{-1}\left(\frac{x}{2}\right) + C \)
\( = -\frac{1}{3} \tan^{-1}x + \frac{2}{3} \tan^{-1}\left(\frac{x}{2}\right) + C \)