CBSE Class 12 Mathematics HOTs Integrals Set 01

Question. Find: \(\int \frac{x^3}{x^4 + 3x^2 + 2} dx\)
Answer: Let \(I = \int \frac{x^3}{x^4 + 3x^2 + 2} dx\)
Put \(x^2 = t \Rightarrow 2x \, dx = dt \Rightarrow x \, dx = \frac{1}{2} dt\)
\(\therefore I = \frac{1}{2} \int \frac{t}{t^2 + 3t + 2} dt = \frac{1}{2} \int \frac{t}{(t+2)(t+1)} dt\)
Let \(\frac{t}{(t+2)(t+1)} = \frac{A}{t+2} + \frac{B}{t+1}\)
\(\Rightarrow t = A(t+1) + B(t+2)\)
Putting \(t = -1, -2\) in it, we get \(A = 2, B = -1\).
\(\therefore \frac{t}{(t+2)(t+1)} = \frac{2}{t+2} - \frac{1}{t+1}\)
\(\Rightarrow I = \frac{1}{2} \int \left[ \frac{2}{t+2} - \frac{1}{t+1} \right] dt\)
\(= \frac{1}{2} \left[ 2 \log|t+2| - \log|t+1| \right] + C\)
\(= \frac{1}{2} \left[ 2 \log|x^2+2| - \log|x^2+1| \right] + C\)

Question. Evaluate: \(\int \frac{2x^2 + 1}{x^2(x^2 + 4)} dx\)
Answer: Let \(I = \int \frac{2x^2 + 1}{x^2(x^2 + 4)} dx\)
Let \(x^2 = y\), then \(\frac{2x^2+1}{x^2(x^2+4)} = \frac{2y+1}{y(y+4)}\)
Let \(\frac{2y+1}{y(y+4)} = \frac{A}{y} + \frac{B}{y+4}\)
\(\Rightarrow 2y+1 = A(y+4) + By\)
Putting \(y = 0, -4\) in it, we get \(A = \frac{1}{4}\) and \(B = \frac{7}{4}\).
\(\therefore \frac{2y+1}{y(y+4)} = \frac{1}{4} \cdot \frac{1}{y} + \frac{7}{4} \cdot \frac{1}{y+4}\)
\(\Rightarrow \frac{2x^2+1}{x^2(x^2+4)} = \frac{1}{4x^2} + \frac{7}{4(x^2+4)}\)
Integrating both sides w.r.t. \(x\), we get:
\(I = \frac{1}{4} \int x^{-2} dx + \frac{7}{4} \int \frac{dx}{x^2+2^2}\)
\(= -\frac{1}{4x} + \frac{7}{8} \tan^{-1}\left(\frac{x}{2}\right) + C\)

Question. Evaluate: \(\int \frac{x^2 + 1}{(x^2 + 4)(x^2 + 25)} dx\)
Answer: Let \(I = \int \frac{x^2 + 1}{(x^2 + 4)(x^2 + 25)} dx\)
Put \(x^2 = y\). Then, \(\frac{x^2+1}{(x^2+4)(x^2+25)} = \frac{y+1}{(y+4)(y+25)}\)
Let \(\frac{y+1}{(y+4)(y+25)} = \frac{A}{y+4} + \frac{B}{y+25}\) ... (1)
\(\Rightarrow y+1 = A(y+25) + B(y+4)\) ... (2)
Putting \(y = -4\) and \(y = -25\) successively in (2), we get \(A = -\frac{1}{7}\) and \(B = \frac{8}{7}\).
Substituting the values of \(A\) and \(B\) in (1), we get:
\(\frac{y+1}{(y+4)(y+25)} = \frac{-1/7}{y+4} + \frac{8/7}{y+25}\)
\(\Rightarrow \frac{x^2+1}{(x^2+4)(x^2+25)} = -\frac{1}{7(x^2+4)} + \frac{8}{7(x^2+25)}\)
\(\therefore I = -\frac{1}{7} \int \frac{1}{x^2+2^2} dx + \frac{8}{7} \int \frac{1}{x^2+5^2} dx\)
\(= -\frac{1}{14} \tan^{-1}\left(\frac{x}{2}\right) + \frac{8}{35} \tan^{-1}\left(\frac{x}{5}\right) + C\)

Question. Evaluate: \(\int \frac{dx}{x(x^5 + 3)}\)
Answer: Let \(I = \int \frac{dx}{x(x^5 + 3)}\)
Put \(x^5 + 3 = t \Rightarrow 5x^4 dx = dt \Rightarrow dx = \frac{dt}{5x^4}\)
\(\therefore I = \int \frac{dt}{5x^5 \cdot t} = \frac{1}{5} \int \frac{dt}{t(t-3)}\)
\(= \frac{1}{5} \int \frac{1}{3} \left[ \frac{1}{t-3} - \frac{1}{t} \right] dt\)
\(= \frac{1}{15} [ \log|t-3| - \log|t| ] + C\)
\(= \frac{1}{15} \log \left| \frac{t-3}{t} \right| + C\)
\(= \frac{1}{15} \log \left| \frac{x^5}{x^5+3} \right| + C\)

Question. Evaluate: \(\int \frac{dx}{x(x^3 + 8)}\)
Answer: Let \(I = \int \frac{dx}{x(x^3 + 8)}\)
Put \(x^3 + 8 = t \Rightarrow 3x^2 dx = dt \Rightarrow dx = \frac{dt}{3x^2}\)
\(\therefore I = \int \frac{dt}{3x^3 \cdot t} = \frac{1}{3} \int \frac{dt}{t(t-8)}\)
\(= \frac{1}{3} \int \frac{1}{8} \left[ \frac{1}{t-8} - \frac{1}{t} \right] dt\)
\(= \frac{1}{24} [ \log|t-8| - \log|t| ] + C\)
\(= \frac{1}{24} \log \left| \frac{t-8}{t} \right| + C\)
\(= \frac{1}{24} \log \left| \frac{x^3}{x^3+8} \right| + C\)

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

Question. Evaluate: \(\int \frac{3x + 1}{(x+1)^2(x+3)} dx\)
Answer: Let \(I = \int \frac{3x + 1}{(x+1)^2(x+3)} dx\)
Let \(\frac{3x+1}{(x+1)^2(x+3)} = \frac{A}{x+1} + \frac{B}{(x+1)^2} + \frac{C}{x+3}\) ... (1)
\(\Rightarrow 3x + 1 = A(x+1)(x+3) + B(x+3) + C(x+1)^2\) ... (2)
Putting \(x = -1, -3, 0\) in (2), we get:
\(B = -1, C = -2, A = 2\)
\(\therefore\) From (1),
\(\frac{3x+1}{(x+1)^2(x+3)} = \frac{2}{x+1} - \frac{1}{(x+1)^2} - \frac{2}{x+3}\)
Integrating both sides w.r.t. \(x\), we get:
\(I = \int \frac{2}{x+1} dx - \int \frac{1}{(x+1)^2} dx - \int \frac{2}{x+3} dx\)
\(= 2\log|x+1| + \frac{1}{x+1} - 2\log|x+3| + C\)
\(= 2\log\left|\frac{x+1}{x+3}\right| + \frac{1}{x+1} + C\)

Question. Evaluate: \(\int \frac{3x + 5}{x^3 - x^2 - x + 1} dx\)
Answer: Let \(I = \int \frac{3x + 5}{x^3 - x^2 - x + 1} dx\)
Here, \(x^3 - x^2 - x + 1 = x^2(x - 1) - 1(x - 1) = (x^2 - 1)(x - 1) = (x - 1)^2(x + 1)\)
Let \(\frac{3x+5}{(x-1)^2(x+1)} = \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{C}{x+1}\) ... (1)
\(\Rightarrow 3x+5 = A(x-1)(x+1) + B(x+1) + C(x-1)^2\) ... (2)
Putting \(x = 1, -1, 0\) in (2), we get:
\(B = 4; C = \frac{1}{2}; A = -\frac{1}{2}\)
From (1),
\(\frac{3x+5}{(x-1)^2(x+1)} = -\frac{1}{2(x-1)} + \frac{4}{(x-1)^2} + \frac{1}{2(x+1)}\)
Integrating, we get:
\(I = -\frac{1}{2}\log|x-1| - \frac{4}{x-1} + \frac{1}{2}\log|x+1| + C\)
\(= \frac{1}{2}\log\left|\frac{x+1}{x-1}\right| - \frac{4}{x-1} + C\)

Question. Evaluate: \(\int \frac{8}{(x+2)(x^2+4)} dx\)
Answer: Let \(I = \int \frac{8}{(x+2)(x^2+4)} dx\)
Let \(\frac{8}{(x+2)(x^2+4)} = \frac{A}{x+2} + \frac{Bx + C}{x^2+4}\) ... (1)
\(\Rightarrow 8 = A(x^2+4) + (Bx+C)(x+2)\) ... (2)
Putting \(x = -2, 0, 1\) in (2), we get:
\(A = 1; C = 2; B = -1\)
From (1), we get:
\(\frac{8}{(x+2)(x^2+4)} = \frac{1}{x+2} + \frac{-x+2}{x^2+4}\)
Integrating, we get:
\(I = \int \frac{1}{x+2} dx + \int \frac{-x+2}{x^2+4} dx\)
\(= \int \frac{1}{x+2} dx - \frac{1}{2} \int \frac{2x}{x^2+4} dx + 2 \int \frac{1}{x^2+2^2} dx\)
\(= \log|x+2| - \frac{1}{2}\log(x^2+4) + \tan^{-1}\left(\frac{x}{2}\right) + C\)
\(= \log\left|\frac{x+2}{\sqrt{x^2+4}}\right| + \tan^{-1}\left(\frac{x}{2}\right) + C\)

Question. Evaluate: \(\int \frac{2}{(1-x)(1+x^2)} dx\)
Answer: Let \(I = \int \frac{2}{(1-x)(1+x^2)} dx\)
Let \(\frac{2}{(1-x)(1+x^2)} = \frac{A}{1-x} + \frac{Bx+C}{1+x^2}\)
\(\Rightarrow 2 = A(1+x^2) + (Bx+C)(1-x)\)
\(\Rightarrow 2 = (A-B)x^2 + (B-C)x + (A+C)\)
Comparing coefficients of \(x^2\), \(x\) and constant terms, we get:
\(A = B = C = 1\)
Hence, \(\frac{2}{(1-x)(1+x^2)} = \frac{1}{1-x} + \frac{x+1}{1+x^2}\)
\(\therefore I = \int \frac{1}{1-x} dx + \int \frac{x+1}{1+x^2} dx\)
\(= -\log|1-x| + \frac{1}{2}\log(1+x^2) + \tan^{-1}x + C\)

Question. Evaluate: \(\int \frac{2x}{(x^2+1)(x^2+3)} dx\)
Answer: Let \(I = \int \frac{2x}{(x^2+1)(x^2+3)} dx\)
Put \(x^2 = y \Rightarrow 2x \, dx = dy\)
\(\therefore I = \int \frac{dy}{(y+1)(y+3)}\)
We write, \(\frac{1}{(y+1)(y+3)} = \frac{A}{y+1} + \frac{B}{y+3}\)
\(\Rightarrow 1 = A(y+3) + B(y+1)\) ... (1)
Putting \(y = -1\) in (1), we get \(A = \frac{1}{2}\).
Putting \(y = -3\) in (1), we get \(B = -\frac{1}{2}\).
\(\therefore I = \frac{1}{2} \int \frac{dy}{y+1} - \frac{1}{2} \int \frac{dy}{y+3}\)
\(= \frac{1}{2}\log|y+1| - \frac{1}{2}\log|y+3| + C\)
\(= \frac{1}{2}\log\left|\frac{x^2+1}{x^2+3}\right| + C\)

Question. Evaluate: \(\int \frac{1-x^2}{x(1-2x)} dx\)
Answer: Since \(\frac{1-x^2}{x(1-2x)} = \frac{x^2-1}{2x^2-x}\) is an improper fraction, we convert it into a proper fraction by division:
\(\frac{x^2-1}{2x^2-x} = \frac{1}{2} + \frac{\frac{1}{2}x - 1}{2x^2-x}\)
\(\Rightarrow I = \int \frac{1}{2} dx + \frac{1}{2} \int \frac{x-2}{x(2x-1)} dx\)
Let \(\frac{x-2}{x(2x-1)} = \frac{A}{x} + \frac{B}{2x-1}\)
\(\Rightarrow x-2 = A(2x-1) + Bx\)
Putting \(x = 0\) in it, we get \(A = 2\).
Putting \(x = \frac{1}{2}\) in it, we get \(B = -3\).
\(\therefore \frac{x-2}{x(2x-1)} = \frac{2}{x} - \frac{3}{2x-1}\)
\(\Rightarrow I = \frac{1}{2}x + \frac{1}{2} [ 2\log|x| - \frac{3}{2}\log|2x-1| ] + C\)
\(= \frac{1}{2}x + \log|x| - \frac{3}{4}\log|2x-1| + C\)

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

Question. Evaluate: \(\int \frac{dx}{\sin x - \sin 2x}\)
Answer: Let \(I = \int \frac{dx}{\sin x(1 - 2\cos x)} = \int \frac{\sin x \, dx}{\sin^2 x(1 - 2\cos x)} = \int \frac{\sin x \, dx}{(1 - \cos^2 x)(1 - 2\cos x)}\)
Put \(\cos x = t \Rightarrow -\sin x \, dx = dt\)
\(\therefore I = \int \frac{-dt}{(1-t^2)(1-2t)} = \int \frac{dt}{(t-1)(t+1)(2t-1)}\)
Using partial fractions:
\(\frac{1}{(t-1)(t+1)(2t-1)} = \frac{A}{t-1} + \frac{B}{t+1} + \frac{C}{2t-1}\)
\(\Rightarrow 1 = A(t+1)(2t-1) + B(t-1)(2t-1) + C(t-1)(t+1)\)
Putting \(t = 1 \Rightarrow 1 = A(2)(1) \Rightarrow A = \frac{1}{2}\).
Putting \(t = -1 \Rightarrow 1 = B(-2)(-3) \Rightarrow B = \frac{1}{6}\).
Putting \(t = \frac{1}{2} \Rightarrow 1 = C(-\frac{1}{2})(\frac{3}{2}) \Rightarrow C = -\frac{4}{3}\).
\(\therefore I = \frac{1}{2} \int \frac{dt}{t-1} + \frac{1}{6} \int \frac{dt}{t+1} - \frac{4}{3} \int \frac{dt}{2t-1}\)
\(= \frac{1}{2}\log|t-1| + \frac{1}{6}\log|t+1| - \frac{2}{3}\log|2t-1| + C\)
\(= \frac{1}{2}\log|1-\cos x| + \frac{1}{6}\log|1+\cos x| - \frac{2}{3}\log|2\cos x-1| + C\)

Question. Evaluate: \(\int \frac{\sin x}{(1-\cos x)(2-\cos x)} dx\)
Answer: Let \(I = \int \frac{\sin x}{(1-\cos x)(2-\cos x)} dx\)
Put \(\cos x = t \Rightarrow -\sin x \, dx = dt\)
\(\therefore I = \int \frac{-dt}{(1-t)(2-t)}\)
We write, \(\frac{-1}{(1-t)(2-t)} = \frac{A}{1-t} + \frac{B}{2-t}\)
\(\Rightarrow -1 = A(2-t) + B(1-t)\) ... (1)
Putting \(t = 1\) in (1), we get \(A = -1\).
Putting \(t = 2\) in (1), we get \(B = 1\).
\(\therefore I = \int \left[ \frac{-1}{1-t} + \frac{1}{2-t} \right] dt\)
\(= \log|1-t| - \log|2-t| + C\)
\(= \log \left| \frac{1-\cos x}{2-\cos x} \right| + C\)

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

Question. Evaluate: \(\int \frac{2x+1}{(x+2)(x-3)} dx\)
Answer: Let \(I = \int \frac{2x+1}{(x+2)(x-3)} dx\)
Let \(\frac{2x+1}{(x+2)(x-3)} = \frac{A}{x+2} + \frac{B}{x-3}\)
\(\Rightarrow 2x+1 = A(x-3) + B(x+2)\) ... (1)
Putting \(x = -2\) in (1), we get \(-3 = A(-5) \Rightarrow A = \frac{3}{5}\).
Putting \(x = 3\) in (1), we get \(7 = B(5) \Rightarrow B = \frac{7}{5}\).
\(\therefore I = \frac{3}{5}\log|x+2| + \frac{7}{5}\log|x-3| + C\)

Question. Find: \(\int \frac{x^2+x+1}{(x+1)^2(x+2)} dx\)
Answer: Let \(I = \int \frac{x^2+x+1}{(x+1)^2(x+2)} dx\)
Let \(\frac{x^2+x+1}{(x+1)^2(x+2)} = \frac{A}{x+1} + \frac{B}{(x+1)^2} + \frac{C}{x+2}\) ... (1)
\(\Rightarrow x^2+x+1 = A(x+1)(x+2) + B(x+2) + C(x+1)^2\)
Putting \(x = -1, -2, 0\), we get:
\(B = 1; C = 3; A = -2\)
From (1), we get:
\(\frac{x^2+x+1}{(x+1)^2(x+2)} = \frac{-2}{x+1} + \frac{1}{(x+1)^2} + \frac{3}{x+2}\)
Integrating both sides:
\(I = -2\log|x+1| - \frac{1}{x+1} + 3\log|x+2| + C\)

Question. Evaluate: \(\int \frac{x^2+1}{(x-1)^2(x+3)} dx\)
Answer: Let \(I = \int \frac{x^2+1}{(x-1)^2(x+3)} dx\)
Let \(\frac{x^2+1}{(x-1)^2(x+3)} = \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{C}{x+3}\)
\(\Rightarrow x^2+1 = A(x-1)(x+3) + B(x+3) + C(x-1)^2\) ... (1)
Putting \(x = 1\) in (1), we get \(B = \frac{1}{2}\).
Putting \(x = -3\) in (1), we get \(C = \frac{5}{8}\).
Putting \(x = 0\) in (1), we get \(A = \frac{3}{8}\).
\(\therefore \frac{x^2+1}{(x-1)^2(x+3)} = \frac{3}{8(x-1)} + \frac{1}{2(x-1)^2} + \frac{5}{8(x+3)}\)
Integrating both sides, we get:
\(I = \frac{3}{8}\log|x-1| - \frac{1}{2(x-1)} + \frac{5}{8}\log|x+3| + C\)

Question. Given \(\int e^x(\tan x + 1)\sec x \, dx = e^x f(x) + c\). Write \(f(x)\) satisfying above.
Answer: Given: \(\int e^x(\tan x + 1)\sec x \, dx = e^x f(x) + C\) ... (1)
L.H.S. \(= \int e^x(\sec x + \sec x \tan x) \, dx\)
Using the standard result \(\int e^x(g(x) + g'(x)) \, dx = e^x g(x) + C\), let \(g(x) = \sec x \Rightarrow g'(x) = \sec x \tan x\).
\(\therefore \text{L.H.S.} = e^x \sec x + C\)
On comparing with (1), we get \(f(x) = \sec x\).

Question. Evaluate: \(\int x \log 2x \, dx\)
Answer: Let \(I = \int x \log 2x \, dx\)
Integrating by parts, taking \(\log 2x\) as the first function:
\(I = \log 2x \int x \, dx - \int \left[ \frac{d}{dx}(\log 2x) \int x \, dx \right] dx\)
\(= \log 2x \cdot \frac{x^2}{2} - \int \left[ \frac{1}{x} \cdot \frac{x^2}{2} \right] dx + C\)
\(= \frac{x^2}{2}\log 2x - \frac{1}{2} \int x \, dx + C\)
\(= \frac{x^2}{2}\log 2x - \frac{x^2}{4} + C\)

Question. Find: \(\int (3x+1)\sqrt{4-3x-2x^2} \, dx\)
Answer: Let \(I = \int (3x+1)\sqrt{4-3x-2x^2} \, dx\)
Let \(3x+1 = \lambda \frac{d}{dx}(4-3x-2x^2) + \mu\)
\(\Rightarrow 3x+1 = \lambda(-3-4x) + \mu\)
On comparing, we get \(\lambda = -\frac{3}{4}\) and \(\mu = -\frac{5}{4}\).
\(\therefore I = \int \left[ -\frac{3}{4}(-3-4x) - \frac{5}{4} \right] \sqrt{4-3x-2x^2} \, dx\)
\(= -\frac{3}{4} \int (-3-4x)\sqrt{4-3x-2x^2} \, dx - \frac{5}{4} \int \sqrt{4-3x-2x^2} \, dx\)
For the first integral, put \(4-3x-2x^2 = t \Rightarrow (-3-4x)dx = dt\):
\(I_1 = \int \sqrt{t} \, dt = \frac{2}{3} t^{3/2} = \frac{2}{3}(4-3x-2x^2)^{3/2}\)
For the second integral:
\(\int \sqrt{4-3x-2x^2} \, dx = \sqrt{2} \int \sqrt{2 - \frac{3}{2}x - x^2} \, dx = \sqrt{2} \int \sqrt{\frac{41}{16} - \left(x+\frac{3}{4}\right)^2} \, dx\)
\(= \sqrt{2} \left[ \frac{x+\frac{3}{4}}{2}\sqrt{\frac{41}{16} - \left(x+\frac{3}{4}\right)^2} + \frac{41}{32}\sin^{-1}\left(\frac{4x+3}{\sqrt{41}}\right) \right]\)
Combining these parts, we obtain:
\(I = -\frac{1}{2}(4-3x-2x^2)^{3/2} - \frac{5}{8\sqrt{2}}\left(x+\frac{3}{4}\right)\sqrt{4-3x-2x^2} - \frac{205}{128\sqrt{2}}\sin^{-1}\left(\frac{4x+3}{\sqrt{41}}\right) + C\)

Question. Find: \(\int \frac{x \sin^{-1} x}{\sqrt{1-x^2}} \, dx\)
Answer: Let \(I = \int \sin^{-1} x \cdot \left(\frac{x}{\sqrt{1-x^2}}\right) dx\)
Integrating by parts, taking \(\sin^{-1} x\) as the first function:
Since \(\int \frac{x}{\sqrt{1-x^2}} dx = -\sqrt{1-x^2}\):
\(I = \sin^{-1} x \left(-\sqrt{1-x^2}\right) - \int \left[ \frac{1}{\sqrt{1-x^2}} \left(-\sqrt{1-x^2}\right) \right] dx\)
\(= -\sqrt{1-x^2}\sin^{-1} x + \int 1 \, dx\)
\(= -\sqrt{1-x^2}\sin^{-1} x + x + C\)

Question. Find: \(\int (2x+5)\sqrt{10-4x-3x^2} \, dx\)
Answer: Let \(I = \int (2x+5)\sqrt{10-4x-3x^2} \, dx\)
Let \(2x+5 = \lambda \frac{d}{dx}(10-4x-3x^2) + \mu\)
\(\Rightarrow 2x+5 = \lambda(-4-6x) + \mu\)
On comparing, we get \(\lambda = -\frac{1}{3}\) and \(\mu = \frac{11}{3}\).
\(\therefore I = -\frac{1}{3} \int (-4-6x)\sqrt{10-4x-3x^2} \, dx + \frac{11}{3} \int \sqrt{10-4x-3x^2} \, dx\)
For the first integral, put \(10-4x-3x^2 = t \Rightarrow (-4-6x)dx = dt\):
\(I_1 = \int \sqrt{t} \, dt = \frac{2}{3} t^{3/2} = \frac{2}{3}(10-4x-3x^2)^{3/2}\)
For the second integral:
\(\int \sqrt{10-4x-3x^2} \, dx = \sqrt{3} \int \sqrt{\frac{10}{3} - \frac{4}{3}x - x^2} \, dx = \sqrt{3} \int \sqrt{\frac{34}{9} - \left(x+\frac{2}{3}\right)^2} \, dx\)
\(= \sqrt{3} \left[ \frac{x+\frac{2}{3}}{2}\sqrt{\frac{34}{9} - \left(x+\frac{2}{3}\right)^2} + \frac{34}{18}\sin^{-1}\left(\frac{3x+2}{\sqrt{34}}\right) \right]\)
Combining these parts, we get:
\(I = -\frac{2}{9}(10-4x-3x^2)^{3/2} + \frac{11}{6}\left(x+\frac{2}{3}\right)\sqrt{10-4x-3x^2} + \frac{187}{18\sqrt{3}}\sin^{-1}\left(\frac{3x+2}{\sqrt{34}}\right) + C\)

Question. Find: \(\int (x+3)\sqrt{3-4x-x^2} \, dx\)
Answer: Let \(I = \int (x+3)\sqrt{3-4x-x^2} \, dx\)
\(= \int \left[ -\frac{1}{2}(-4-2x) + 1 \right] \sqrt{3-4x-x^2} \, dx\)
\(= -\frac{1}{2} \int (-4-2x)\sqrt{3-4x-x^2} \, dx + \int \sqrt{3-4x-x^2} \, dx\)
\(= I_1 + I_2\)
For \(I_1\), put \(3-4x-x^2 = t \Rightarrow (-4-2x)dx = dt\):
\(I_1 = -\frac{1}{2} \int \sqrt{t} \, dt = -\frac{1}{3} t^{3/2} = -\frac{1}{3}(3-4x-x^2)^{3/2}\)
For \(I_2\):
\(I_2 = \int \sqrt{7 - (x+2)^2} \, dx = \frac{x+2}{2}\sqrt{3-4x-x^2} + \frac{7}{2}\sin^{-1}\left(\frac{x+2}{\sqrt{7}}\right)\)
\(\therefore I = -\frac{1}{3}(3-4x-x^2)^{3/2} + \frac{x+2}{2}\sqrt{3-4x-x^2} + \frac{7}{2}\sin^{-1}\left(\frac{x+2}{\sqrt{7}}\right) + C\)

Question. Integrate the following w.r.t. x : \(\frac{x^2-3x+1}{\sqrt{1-x^2}}\)
Answer: Let \(I = \int \frac{x^2-3x+1}{\sqrt{1-x^2}} \, dx\)
\(= \int \frac{-(1-x^2) - 3x + 2}{\sqrt{1-x^2}} \, dx\)
\(= -\int \sqrt{1-x^2} \, dx - 3\int \frac{x}{\sqrt{1-x^2}} \, dx + 2\int \frac{1}{\sqrt{1-x^2}} \, dx\)
\(= -\left[ \frac{x}{2}\sqrt{1-x^2} + \frac{1}{2}\sin^{-1} x \right] + 3\sqrt{1-x^2} + 2\sin^{-1} x + C\)
\(= \left(3-\frac{x}{2}\right)\sqrt{1-x^2} + \frac{3}{2}\sin^{-1} x + C\)

Question. Evaluate: \(\int (3-2x)\sqrt{2+x-x^2} \, dx\)
Answer: Let \(I = \int (3-2x)\sqrt{2+x-x^2} \, dx\)
\(= \int (1-2x+2)\sqrt{2+x-x^2} \, dx\)
\(= \int (1-2x)\sqrt{2+x-x^2} \, dx + 2 \int \sqrt{2+x-x^2} \, dx\)
\(= I_1 + 2I_2\)
For \(I_1\), let \(2+x-x^2 = t \Rightarrow (1-2x)dx = dt\):
\(I_1 = \int \sqrt{t} \, dt = \frac{2}{3} t^{3/2} = \frac{2}{3}(2+x-x^2)^{3/2}\)
For \(I_2\):
\(I_2 = \int \sqrt{\frac{9}{4} - \left(x-\frac{1}{2}\right)^2} \, dx = \frac{x-\frac{1}{2}}{2}\sqrt{2+x-x^2} + \frac{9}{8}\sin^{-1}\left(\frac{2x-1}{3}\right)\)
\(\therefore I = \frac{2}{3}(2+x-x^2)^{3/2} + \left(x-\frac{1}{2}\right)\sqrt{2+x-x^2} + \frac{9}{4}\sin^{-1}\left(\frac{2x-1}{3}\right) + C\)

Question. Find: \(\int \frac{\log x}{(x+1)^2} \, dx\)
Answer: Let \(I = \int \log x \cdot (x+1)^{-2} \, dx\)
Integrating by parts, taking \(\log x\) as the first function:
\(I = \log x \cdot \frac{(x+1)^{-1}}{-1} - \int \left[ \frac{1}{x} \cdot \frac{(x+1)^{-1}}{-1} \right] dx\)
\(= -\frac{\log x}{x+1} + \int \frac{1}{x(x+1)} dx\)
\(= -\frac{\log x}{x+1} + \int \left[ \frac{1}{x} - \frac{1}{x+1} \right] dx\)
\(= -\frac{\log x}{x+1} + \log|x| - \log|x+1| + C\)
\(= -\frac{\log x}{x+1} + \log\left|\frac{x}{x+1}\right| + C\)

Question. Evaluate: \(\int e^{2x} \cdot \sin(3x+1) \, dx\)
Answer: Let \(I = \int e^{2x}\sin(3x+1) \, dx\)
Integrating by parts, taking \(\sin(3x+1)\) as the first function:
\(I = \sin(3x+1) \cdot \frac{e^{2x}}{2} - \int 3\cos(3x+1) \cdot \frac{e^{2x}}{2} \, dx\)
\(= \frac{e^{2x}}{2}\sin(3x+1) - \frac{3}{2} \int e^{2x}\cos(3x+1) \, dx\)
Let \(I_1 = \int e^{2x}\cos(3x+1) \, dx\). Integrating \(I_1\) by parts:
\(I_1 = \cos(3x+1) \cdot \frac{e^{2x}}{2} - \int -3\sin(3x+1) \cdot \frac{e^{2x}}{2} \, dx\)
\(= \frac{e^{2x}}{2}\cos(3x+1) + \frac{3}{2} I\)
Substituting this back into the expression for \(I\):
\(I = \frac{e^{2x}}{2}\sin(3x+1) - \frac{3}{2} \left[ \frac{e^{2x}}{2}\cos(3x+1) + \frac{3}{2} I \right]\)
\(I = \frac{e^{2x}}{2}\sin(3x+1) - \frac{3e^{2x}}{4}\cos(3x+1) - \frac{9}{4} I\)
\(\Rightarrow \frac{13}{4} I = e^{2x} \left[ \frac{1}{2}\sin(3x+1) - \frac{3}{4}\cos(3x+1) \right]\)
\(\Rightarrow I = \frac{e^{2x}}{13} [ 2\sin(3x+1) - 3\cos(3x+1) ] + C\)

Question. Find: \(\int \frac{(x^2+1)e^x}{(x+1)^2} \, dx\)
Answer: Let \(I = \int e^x \left[ \frac{x^2+1}{(x+1)^2} \right] dx = \int e^x \left[ \frac{x^2-1+2}{(x+1)^2} \right] dx\)
\(= \int e^x \left[ \frac{x-1}{x+1} + \frac{2}{(x+1)^2} \right] dx\)
Let \(f(x) = \frac{x-1}{x+1} \Rightarrow f'(x) = \frac{(x+1)(1) - (x-1)(1)}{(x+1)^2} = \frac{2}{(x+1)^2}\)
Using the standard rule \(\int e^x [f(x) + f'(x)] \, dx = e^x f(x) + C\):
\(I = e^x \left( \frac{x-1}{x+1} \right) + C\)

Question. Evaluate: \(\int (x-3)\sqrt{x^2+3x-18} \, dx\)
Answer: Let \(I = \int (x-3)\sqrt{x^2+3x-18} \, dx\)
\(= \int \left[ \frac{1}{2}(2x+3) - \frac{9}{2} \right] \sqrt{x^2+3x-18} \, dx\)
\(= \frac{1}{2} \int (2x+3)\sqrt{x^2+3x-18} \, dx - \frac{9}{2} \int \sqrt{x^2+3x-18} \, dx\)
\(= I_1 - \frac{9}{2} I_2\)
For \(I_1\), let \(x^2+3x-18 = t \Rightarrow (2x+3)dx = dt\):
\(I_1 = \frac{1}{2}\int \sqrt{t} \, dt = \frac{1}{3} t^{3/2} = \frac{1}{3}(x^2+3x-18)^{3/2}\)
For \(I_2\):
\(I_2 = \int \sqrt{\left(x+\frac{3}{2}\right)^2 - \frac{81}{4}} \, dx = \frac{2x+3}{4}\sqrt{x^2+3x-18} - \frac{81}{8}\log\left| \left(x+\frac{3}{2}\right) + \sqrt{x^2+3x-18} \right|\)
\(\therefore I = \frac{1}{3}(x^2+3x-18)^{3/2} - \frac{9}{8}(2x+3)\sqrt{x^2+3x-18} + \frac{729}{16}\log\left| \left(x+\frac{3}{2}\right) + \sqrt{x^2+3x-18} \right| + C\)

Question. Evaluate: \(\int \frac{x \cos^{-1} x}{\sqrt{1-x^2}} \, dx\)
Answer: Let \(I = \int \frac{x \cos^{-1} x}{\sqrt{1-x^2}} \, dx\)
Put \(\cos^{-1} x = \theta \Rightarrow x = \cos\theta \Rightarrow dx = -\sin\theta \, d\theta\)
\(\therefore I = \int \frac{\cos\theta \cdot \theta}{\sqrt{1-\cos^2\theta}} (-\sin\theta) d\theta = -\int \theta \cos\theta \, d\theta\)
Integrating by parts:
\(I = -\left[ \theta \sin\theta - \int \sin\theta \, d\theta \right] = -\theta \sin\theta - \cos\theta + C\)
Converting back to \(x\):
\(I = -\sqrt{1-x^2}\cos^{-1} x - x + C\)