\(\sin (a x+b) \cos (a x+b)=\frac{2 \sin (a x+b) \cos (a x+b)}{2}=\frac{\sin 2(a x+b)}{2} \)
\(\text { Let } 2(a x+b)=t \)
\(\therefore 2 a d x=d t \)
\(\Rightarrow \int \frac{\sin 2(a x+b)}{2} d x =\frac{1}{2} \int \frac{\sin t d t}{2 a} \)
\(=\frac{1}{4 a}[-\cos t]+\mathrm{C} \)
\(=\frac{-1}{4 a} \cos 2(a x+b)+\mathrm{C}\)

\(I=\int { \frac { 1 }{ x+x\log { x } } } dx\)
\(=\int { \frac { 1 }{ x(1+\log { x } ) } dx } \)
Put 1+log x=t so that \(\frac { 1 }{ x } dx=dt\)
\(\therefore \ I=\int { \frac { 1 }{ t } dt } =\log { \left| t \right| } +C\)
\(=\log { \left| 1+\log { x } \right| } +C.\)

Let \(1+{ x }^{ 2 }=t\)so that
2x dx=dt.
\(\therefore I=\int { \frac { dt }{ t } = } \log { |t| } +C\)
\(=\log { \left| 1+{ x }^{ 2 } \right| } +C\)
\(= \log { (1+{ x }^{ 2 }) } +C\)
\(\left[ \because { x }^{ 2 }\ge 0\Rightarrow 1+{ x }^{ 2 }>0\Rightarrow \left| 1+{ x }^{ 2 } \right| =1+{ x }^{ 2 } \right] \)

Part (A) is the correct answer.
Reason: We have: \(\frac { d }{ dx } f\left( x \right) ={ 4x }^{ 3 }-\frac { 3 }{ { x }^{ 4 } } \)
Integrating, \(f\left( x \right) =\int { \left( { 4x }^{ 3 }-\frac { 3 }{ { x }^{ 4 } } \right) } dx\)
\(=4\int { { x }^{ 3 } } dx-3\quad \int { { x }^{ -4 } } dx\)
\(=4.\frac { { x }^{ 4 } }{ 4 } -3\frac { { x }^{ -4+1 } }{ -4+1 } +3\)
\( ={ x }^{ 4 }-\frac { { 3x }^{ -3 } }{ -3 } +C\)
\(={ x }^{ 4 }+\frac { 1 }{ { x }^{ 3 } } +C\)
But \(f(2)=0\Rightarrow { (2) }^{ 4 }+\frac { 1 }{ { (2) }^{ 3 } } +C=0\)
\(\Rightarrow 16+\frac { 1 }{ 8 } +C=0\)
\( \Rightarrow C=-16-\frac { 1 }{ 8 } =-\frac { 129 }{ 8 } \)
Putting in (1), f(x) = \({ x }^{ 4 }+\frac { 1 }{ { x }^{ 2 } } -\frac { 129 }{ 8 } \)

\(\int\left(2 x-3 \cos x+e^{x}\right) d x \)
\(=2 \int x d x-3 \int \cos x d x+\int e^{x} d x \)
\(=\frac{2 x^{2}}{2}-3(\sin x)+e^{x}+C \)
\(=x^{2}-3 \sin x+e^{x}+C \)

\(\int { \sqrt { x } } \left( { 3x }^{ 2 }+2x+3 \right) dx.\)
\(=3\int { { x }^{ \frac { 5 }{ 2 } } } dx+2\int { { x }^{ \frac { 3 }{ 2 } } } dx+3\int { { x }^{ \frac { 1 }{ 2 } } } dx\)
\(=3\frac { { x }^{ \frac { 5 }{ 2 } +1 } }{ \frac { 5 }{ 2 } +1 } +2\frac { { x }^{ \frac { 3 }{ 2 } +1 } }{ \frac { 3 }{ 2 } +1 } +3\frac { { x }^{ \frac { 1 }{ 2 } +1 } }{ \frac { 1 }{ 2 } +1 } +C\)
\(=3\times \frac { 2 }{ 7 } { x }^{ \frac { 7 }{ 2 } }+2\times \frac { 2 }{ 5 } { x }^{ \frac { 5 }{ 2 } }+3\times \frac { 2 }{ 3 } { x }^{ \frac { 3 }{ 2 } }+C\)
\(=\frac { 6 }{ 7 } { x }^{ \frac { 7 }{ 2 } }+\frac { 4 }{ 5 } { x }^{ \frac { 5 }{ 2 } }+2{ x }^{ \frac { 3 }{ 2 } }+C.\)

The anti derivative of (ax + b)³ is the function of x whose derivative is (ax + b)³
It is known that,
\(\frac { d }{ dx } \left( (ax+{ b) }^{ 3 } \right) =3a(ax+{ b) }^{ 2 }\)
\(\Rightarrow \frac { d }{ dx } \left( \frac { 1 }{ 3a } (ax+{ b) }^{ 3 } \right) =(ax+{ b) }^{ 2 }\)
Hence,  \(\int { (ax+{ b) }^{ 2 } } dx=\frac { 1 }{ 3a } (ax+{ b) }^{ 3 }+C.\)

The anti derivative of e^2x is the function of x whose derivative is e^2x.
It is known that,
\(\frac { d }{ dx } ({ e }^{ 2x })={ 2e }^{ 2x }.\)
\(\Rightarrow \frac { d }{ dx } \left( \frac { 1 }{ 2 } { e }^{ 2x } \right) ={ e }^{ 2x }.\)
\(\int { { e }^{ 2x } } dx=\frac { 1 }{ 2 } { e }^{ 2x }+C.\)
Therefore, the anti derivative of \(e^{2 x} \text { is } \frac{1}{2} e^{2 x} \text { . }\)

\(\text { Let } I=\int \frac{\sin 2 x \cos 2 x}{\sqrt{9-\cos ^{4} 2 x}} d x\)
Put cos² (2x) = t so that 4 sin 2x cos 2x dx = – dt
\(\text { Therefore } \quad \mathrm{I}=-\frac{1}{4} \int \frac{d t}{\sqrt{9-t^{2}}}=-\frac{1}{4} \sin ^{-1}\left(\frac{t}{3}\right)+\mathrm{C}=-\frac{1}{4} \sin ^{-1}\left[\frac{1}{3} \cos ^{2} 2 x\right]+\mathrm{C}\)

We have
\(\frac{x^{4}}{(x-1)\left(x^{2}+1\right)} =(x+1)+\frac{1}{x^{3}-x^{2}+x-1} \)
\(=(x+1)+\frac{1}{(x-1)\left(x^{2}+1\right)} \)
\(\text {Now express } \ \frac{1}{(x-1)\left(x^{2}+1\right)}=\frac{\mathrm{A}}{(x-1)}+\frac{\mathrm{B} x+\mathrm{C}}{\left(x^{2}+1\right)}\)
1 = A (x² + 1) + (Bx + C) (x – 1)
= (A + B) x² + (C – B) x + A – C
Equating coefficients on both sides, we get A + B = 0, C – B = 0 and A – C = 1,
which give \( \mathrm{A}=\frac{1}{2}, \mathrm{~B}=\mathrm{C}=-\frac{1}{2} \) Substituting values of A,B and C in (2), we get 
\(\frac{1}{(x-1)\left(x^{2}+1\right)}=\frac{1}{2(x-1)}-\frac{1}{2} \frac{x}{\left(x^{2}+1\right)}-\frac{1}{2\left(x^{2}+1\right)}\)
 Again, substituting (3) in (1), we have 
\(\frac{x^{4}}{(x-1)\left(x^{2}+x+1\right)}=(x+1)+\frac{1}{2(x-1)}-\frac{1}{2} \frac{x}{\left(x^{2}+1\right)}-\frac{1}{2\left(x^{2}+1\right)}\)
\(\text {Therefore}\ \int \frac{x^{4}}{(x-1)\left(x^{2}+x+1\right)} d x=\frac{x^{2}}{2}+x+\frac{1}{2} \log |x-1|-\frac{1}{4} \log \left(x^{2}+1\right)-\frac{1}{2} \tan ^{-1} x+\mathrm{C}\)

\(I=\int { \sqrt { { 2 }^{ 2 }-{ (x-1) }^{ 2 } } } \)
\(Put\quad x+1=t\ so\ that \ dx=dt.\)
\( \therefore I=\int { \sqrt { { 2 }^{ 2 }-t^{ 2 } } } dt|\quad From:\int { \sqrt { { a }^{ 2 }-x^{ 2 } } } dx\)
\(=\frac { t\sqrt { 4-{ t }^{ 2 } } }{ 2 } +\frac { 4 }{ 2 } \sin ^{ -1 }{ \frac { t }{ 2 } }\)
\(=\frac { (x+1)\sqrt { 3-2x-{ x }^{ 2 } } }{ 2 } +2\sin ^{ -1 }{ \frac { x+1 }{ 2 } +c } \)

\(\text { Consider } \frac{x^{2}}{\left(x^{2}+1\right)\left(x^{2}+4\right)} \text { and put } x^{2}=y \text { . }\)
\(\text { Then }\frac{x^{2}}{\left(x^{2}+1\right)\left(x^{2}+4\right)}=\frac{y}{(y+1)(y+4)} \)
\(\text { Write } \quad \frac{y}{(y+1)(y+4)}=\frac{\mathrm{A}}{y+1}+\frac{\mathrm{B}}{y+4}\)
So that y = A (y + 4) + B (y + 1)
Comparing coefficients of y and constant terms on both sides, we get A + B = 1 and 4A + B = 0, which give
\(\mathrm{A}=-\frac{1}{3} \quad \text { and } \quad \mathrm{B}=\frac{4}{3}\)
\(\text { Thus, } \quad \frac{x^{2}}{\left(x^{2}+1\right)\left(x^{2}+4\right)}=-\frac{1}{3\left(x^{2}+1\right)}+\frac{4}{3\left(x^{2}+4\right)}\)
\(\text { Therefore, } \quad \int \frac{x^{2} d x}{\left(x^{2}+1\right)\left(x^{2}+4\right)}=-\frac{1}{3} \int \frac{d x}{x^{2}+1}+\frac{4}{3} \int \frac{d x}{x^{2}+4}\)
\(=-\frac{1}{3} \tan ^{-1} x+\frac{4}{3} \times \frac{1}{2} \tan ^{-1} \frac{x}{2}+\mathrm{C} \)
\(=-\frac{1}{3} \tan ^{-1} x+\frac{2}{3} \tan ^{-1} \frac{x}{2}+\mathrm{C} \)
In the above example, the substitution was made only for the partial fraction part and not for the integration part. Now, we consider an example, where the integration involves a combination of the substitution method and the partial fraction method.

Here the integrand \(\frac { { x }^{ 2 }+1 }{ { x }^{ 2 }-5x+6 } \)is not a proper rational function, so dividing \(({ x }^{ 2 }+1)by{ (x }^{ 2 }-5x+6)\), we devide \(x^2+1 \text { by } x^2-5 x+6\) and find that
\(\frac{x^{2}+1}{x^{2}-5 x+6} =1+\frac{5 x-5}{x^{2}-5 x+6}=1+\frac{5 x-5}{(x-2)(x-3)} \)
\(\frac{5 x-5}{(x-2)(x-3)} =\frac{\mathrm{A}}{x-2}+\frac{\mathrm{B}}{x-3}\)
So that 5x – 5 = A (x – 3) + B (x – 2)
Equating the coefficients of x and constant terms on both sides, we get A + B = 5 and 3A + 2B = 5. Solving these equations, we get A = – 5 and B = 10
\(\text {Thus,} \frac{x^{2}+1}{x^{2}-5 x+6}=1-\frac{5}{x-2}+\frac{10}{x-3}\)
\(\therefore \int \frac{x^{2}+1}{x^{2}-5 x+6} d x=\int d x-5 \int \frac{1}{x-2} d x+10 \int \frac{d x}{x-3}\)
= x – 5 log | x – 2 | + 10 log | x – 3 | + C.

(i) We have
\(\int \frac{x^{3}-1}{x^{2}} d x=\int x d x-\int x^{-2} d x \ \text { (by Property V) }\)
\(=\left(\frac{x^{1+1}}{1+1}+C_{1}\right)-\left(\frac{x^{-2+1}}{-2+1}+C_{2}\right) ; C_{1}, C_{2} \text { are constants of integration }\)
\(=\frac{x^{2}}{2}+C_{1}-\frac{x^{-1}}{-1}-C_{2}=\frac{x^{2}}{2}+\frac{1}{x}+C_{1}-C_{2}\)
\(=\frac{x^{2}}{2}+\frac{1}{x}+\mathrm{C}, \text { where } \mathrm{C}=\mathrm{C}_{1}-\mathrm{C}_{2} \text { is another constant of integration. } \)
(ii) We have
\(\int\left(x^{\frac{2}{3}}+1\right) d x =\int x^{\frac{2}{3}} d x+\int d x \)
\(=\frac{x^{\frac{2}{3}+1}}{\frac{2}{3}+1}+x+C=\frac{3}{5} x^{\frac{5}{3}}+x+C \)
(iii) We have
\(\int\left(x^{\frac{3}{2}}+2 e^{x}-\frac{1}{x}\right) d x =\int x^{\frac{3}{2}} d x+\int 2 e^{x} d x-\int \frac{1}{x} d x \)
\(=\frac{x^{\frac{3}{2}+1}}{\frac{3}{2}+1}+2 e^{x}-\log |x|+\mathrm{C} \)
\(=\frac{2}{5} x^{\frac{5}{2}}+2 e^{x}-\log |x|+\mathrm{C} \) 

(i) We know that  \(\frac { d }{ dx } (\sin { 2x) } =2\cos { 2x } \)
\(\Rightarrow \cos { 2x } =\frac { 1 }{ 2 } \frac { d }{ dx } (\sin { 2x) } =\frac { d }{ dx } (\frac { 1 }{ 2 } \sin { 2x) } \)
Hence, antiderivative of \(\cos { 2x } =\frac { 1 }{ 2 } \sin { 2x } \)
(ii) We know that \(\frac { d }{ dx } ({ x }^{ 3 }+{ x }^{ 4 })={ 3x }^{ 2 }+{ 4x }^{ 3 }.\)
Hence, antiderivative of \({ 3x }^{ 2 }+{ 4x }^{ 3 }+{ x }^{ 3 }+{ x }^{ 4 }\)
(iii) We know that \(\frac { d }{ dx } (\log { x } )=\frac { 1 }{ x } ,x>0\) and \(\frac { d }{ dx } [\log { (-x) } ]=\frac { 1 }{ -x } (-1)=\frac { 1 }{ x } ,x>0.\)
combining, \(\frac { d }{ dx } (\log { 1\quad x1 } )=\frac { 1 }{ x } ,x\neq 0.\)
\(\Rightarrow \int { \frac { 1 }{ x } } dx=\log { 1\quad x1 } .\)
which is one of antiderivative of \(\frac { 1 }{ x } \).