Let \(l=\int \frac{1}{\sin ^4 x+\sin ^2 x \cos ^2 x+\cos ^4 x} d x\)
On dividing numerator and denominator by \(\cos ^4 x\) in RHS, we get
\(I=\int \frac{\sec ^4 x}{\tan ^4 x+\tan ^2 x+1} d x=\int \frac{\left(\sec ^2 x\right)\left(\sec ^2 x\right)}{\tan ^4 x+\tan ^2 x+1} d x\)
Put \(\tan x=t \Rightarrow \sec ^2 x d x=d t\)
and \(\sec ^2 x=1+\tan ^2 x=1+t^2\)
\(\therefore I=\int \frac{1+t^2}{t^4+t^2+1} d t=\int \frac{1+\frac{1}{t^2}}{t^2+\frac{1}{t^2}+1} d t\)
\(\left[\because\right.\) divide numerator and denominator by \(\left.t^2\right]\)
\(=\int \frac{1+\frac{1}{t^2}}{t^2+\frac{1}{t^2}-2+2+1} d t=\int \frac{1+\frac{1}{t^2}}{\left(t-\frac{1}{t}\right)^2+3} d t\)
Again, put \(u=t-\frac{1}{t} \Rightarrow\left(1+\frac{1}{t^2}\right) d t=d u\)
\(\therefore  I=\int \frac{d u}{u^2+(\sqrt{3})^2} \Rightarrow I=\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{u}{\sqrt{3}}\right)+C\)
\( {\left[\because \int \frac{d x}{x^2+a^2}=\frac{1}{a} \tan ^{-1}\left(\frac{x}{a}\right)+C\right] }\)
\(=\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{t-\frac{1}{t}}{\sqrt{3}}\right)+C \quad\left[\because u=t-\frac{1}{t}\right] \)
\(=\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{t^2-1}{\sqrt{3} t}\right)+C\)
\(\therefore \quad I  =\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{\tan ^2 x-1}{\sqrt{3} \tan x}\right)+C\)

Let \(I=\int \frac{\sin x}{\sin ^3 x+\cos ^3 x} d x=\int \frac{\sin x / \cos ^3 x}{\frac{\sin ^3 x}{\cos ^3 x}+\frac{\cos ^3 x}{\cos ^3 x}} d x\)
\(\left[\because\right.\) divide numerator and denominator by\(\left.\cos ^3 x\right]\)
\(=\int \frac{\tan x \sec ^2 x}{\tan ^3 x+1} d x\)
On substituting \(\tan x=t\) and \(\sec ^2 x d x=d t\), we get
\(I =\int \frac{t}{t^3+1} d t=\int \frac{t}{(t+1)\left(t^2-t+1\right)} d t \)
\(=-\frac{1}{3} \int \frac{1}{t+1} d t+\frac{1}{3} \int \frac{t+1}{t^2-t+1} d t \)
\(=-\frac{1}{3} \log |t+1|+\frac{1}{6} \int \frac{(2 t-1)+3}{t^2-t+1} d t\)
\(=-\frac{1}{3} \log |t+1|+\frac{1}{6} \int \frac{2 t-1}{t^2-t+1} d t+\frac{1}{2} \int \frac{1}{t^2-t+1} d t\)
\(=-\frac{1}{3} \log |t+1|+\frac{1}{6} \log \left|t^2-t+1\right|+\frac{1}{2} \int \frac{1}{\left(t-\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2} d t\)
\(\left[\because t^2-t+1=z \Rightarrow(2 t-1) d t=d z\right]\)
\(\left.\therefore \int \frac{2 t-1}{t^2-t+1} d t=\int \frac{d z}{z}=\log |z|=\log \left|t^2-t+1\right|\right]\)
\(=-\frac{1}{3} \log |t+1|+\frac{1}{6} \log \left|t^2-t+1\right|+\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{2 t-1}{\sqrt{3}}\right)\)
\(\left[\because \int \frac{d x}{x^2+a^2}=\frac{1}{a} \tan ^{-1} \frac{x}{a}+C\right]\)
\(=-\frac{1}{3} \log |\tan x+1|+\frac{1}{6} \log \left|\tan ^2 x-\tan x+1\right|+\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{2 \tan x-1}{\sqrt{3}}\right)+C[\because t=\tan x]\)

Let \(I =\int \frac{\sin x-\cos x}{\sqrt{\sin 2 x}} d x=\int \frac{\sin x-\cos x}{\sqrt{1+\sin 2 x-1}} d x\)
\(=\int \frac{\sin x-\cos x}{\sqrt{\sin ^2 x+\cos ^2 x+2 \sin x \cos x-1}} d x \)
\(=\int \frac{\sin x-\cos x}{\sqrt{(\sin x+\cos x)^2-1}} d x\)
Put \(\sin x+\cos x=t \Rightarrow(\cos x-\sin x) d x=d t\)
\(\therefore  I=\int  \frac{-d t}{\sqrt{t^2-1}}=-\log \left|t+\sqrt{t^2-1}\right|+C\)
\( {\left[\because \int \frac{d x}{\sqrt{x^2-a^2}}=\log \left|x+\sqrt{x^2-a^2}\right|+C\right] }\)
\(\Rightarrow I=  -\log \mid(\sin x+\cos x) +\sqrt{(\sin x+\cos x)^2-1} \mid+C\)
\([\because t=\sin x+\cos x] \)
\(=  -\log \mid(\sin x+\cos x) +\sqrt{\sin ^2 x+\cos ^2 x+2 \sin x \cos x-1} \mid+C\)
\(=  -\log |(\sin x+\cos x)+\sqrt{\sin 2 x}|+C\)

Let \(I=\int \frac{x^2+4}{x^4+16} d x\)
On dividing numerator and denominator by \(x^2\), we get
\(I=\int \frac{\left(1+\frac{4}{x^2}\right)}{\left(x^2+\frac{16}{x^2}\right)} d x =\int \frac{\left(1+\frac{4}{x^2}\right)}{x^2+\left(\frac{4}{x}\right)^2-8+8} d x\)
\(=\int \frac{\left(1+\frac{4}{x^2}\right)}{\left(x-\frac{4}{x}\right)^2+8} d x\)
Put \(x-\frac{4}{x}=t \Rightarrow\left(1+\frac{4}{x^2}\right) d x=d t\)
\(\therefore  =\int \frac{d t}{t^2+8}=\int \frac{d t}{t^2+(2 \sqrt{2})^2} \)
\( =\frac{1}{2 \sqrt{2}} \tan ^{-1}\left(\frac{t}{2 \sqrt{2}}\right)+C\)
\({\left[\because \int \frac{d x}{a^2+x^2}=\frac{1}{a} \tan ^{-1}\left(\frac{x}{a}\right)+C\right] }\)
\( =\frac{1}{2 \sqrt{2}} \tan ^{-1}\left(\frac{x-\frac{4}{x}}{2 \sqrt{2}}\right)+C \quad\left[\because t=x-\frac{4}{x}\right]\)
\( =\frac{1}{2 \sqrt{2}} \tan ^{-1}\left(\frac{x-{4}{}}{2 \sqrt{2x}}\right)+C\)

\(\text { Let } I=\int \sin x \sin 2 x \sin 3 x d x\)
\(=\frac{1}{2} \int \sin x(2 \sin 2 x \sin 3 x) d x\)
\(\text { [multiplying numerator and denominator by } 2 \text { ] }\)
\(=\frac{1}{2} \int \sin x[\cos (2 x-3 x)-\cos (2 x+3 x)] d x\)
\([\because 2 \sin A \sin B=\cos (A-B)-\cos (A+B)]\)
\(=\frac{1}{2} \int \sin x[\cos (-x)-\cos 5 x] d x\)
\(=\frac{1}{2} \int \sin x(\cos x-\cos 5 x) d x \quad[\because \cos (-x)=\cos x]\)
\(=\frac{1}{2} \int \sin x \cos x d x-\frac{1}{2} \int \sin x \cos 5 x d x\)
\(=\frac{1}{4} \int 2 \sin x \cos x d x-\frac{1}{4} \int(2 \sin x \cos 5 x) d x\)
[multiplying numerator and denominator by 2]
\(=\frac{1}{4} \int \sin 2 x d x-\frac{1}{4} \int\{\sin (x+5 x)+\sin (x-5 x)\} d x\)
\(\because 2 \sin A \cos A=\sin 2 A \text { and } 2 \sin A \cos B=\sin (A+B)+\sin (A-B)\)
\(=\frac{1}{4} \int \sin 2 x d x-\frac{1}{4} \int[\sin 6 x+\sin (-4 x)] d x\)
\(=\frac{1}{4} \int \sin 2 x d x-\frac{1}{4} \int(\sin 6 x-\sin 4 x) d x\)
\([\because \sin (-\theta)=-\sin \theta]\)
\(=\frac{-1}{4} \cdot \frac{\cos 2 x}{2}-\frac{1}{4}\left[\frac{-\cos 6 x}{6}+\frac{\cos 4 x}{4}\right]+C\) 
\({\left[\because \int \sin a x d x=\frac{-\cos a x}{a}+C\right]}\)
\(=\frac{-\cos 2 x}{8}+\frac{\cos 6 x}{24}-\frac{\cos 4 x}{16}+C\)

Let \(I =\int \frac{3 x+5}{x^3-x^2-x+1} d x=\int \frac{3 x+5}{x^2(x-1)-1(x-1)} d x\)
\(=\int \frac{3 x+5}{(x-1)\left(x^2-1\right)} d x=\int \frac{3 x+5}{(x-1)(x-1)(x+1)} d x \)
\(=\int \frac{3 x+5}{(x-1)^2(x+1)} d x\)
\(\left[\right.\left.\frac{1}{2} \log \left|\frac{x+1}{x-1}\right|-\frac{4}{x-1}+C\right]\)

Let \(I=\int \frac{d x}{x\left(x^5+3\right)}=\int \frac{x^4}{x^5\left(x^5+3\right)} d x\)
[\(\because\) multiplying numerator and denominator by x⁴]
Put \(t=x^5 \Rightarrow d t=5 x^4 d x\)
\(\therefore \quad I=\int \frac{d t}{5 t(t+3)}=\frac{1}{5} \int\left[\frac{1 / 3}{t}-\frac{1 / 3}{t+3}\right] d t\)
[by using partial fraction] 
\(=\frac{1}{5} \int \frac{1}{3}\left[\frac{1}{t}-\frac{1}{t+3}\right] d t \)
\(=\frac{1}{15}[\log |t|-\log |t+3|]+C =\frac{1}{15} \log \left|\frac{t}{t+3}\right|+C\)
\(=\frac{1}{15} \log \left|\frac{x^5}{x^5+3}\right|+C \quad\left[\because t=x^5\right]\)

Let \(l=\int e^{2 x}\left(\frac{1-\sin 2 x}{1-\cos 2 x}\right) d x\)
\(=\int e^{2 x}\left(\frac{1-2 \sin x \cos x}{2 \sin ^2 x}\right) d x\)
\(\because 1-\cos 2 \theta=2 \sin ^2 \theta\)
\(\text { and } \sin 2 \theta=2 \sin \theta \cos \theta\)
\(=\frac{1}{2} \int e^{2 x}\left(\operatorname{cosec}^2 x-2 \cot x\right) d x \)
\(=\frac{1}{2} \int e_1^{2 x} \operatorname{cosec}^2 x d x-\int e^{2 x} \cot x d x \)
\(=\frac{1}{2}\left[e^{2 x} \int \operatorname{cosec}^2 x d x-\int\left\{\frac{d}{d x}\left(e^{2 x}\right) \int \operatorname{cosec}^2 x d x\right\} d x\right] -\int e^{2 x} \cot x d x\)
[by using integration by parts]
\(=\frac{1}{2}\left[-e^{2 x} \cot x+\int 2 e^{2 x} \cot x d x\right]+C-\int e^{2 x} \cot x d x\)
\(=-\frac{e^{2 x}}{2} \cot x+\int e^{2 x} \cot x d x-\int e^{2 x} \cot x d x+C\)
\(=-\frac{e^{2 x}}{2} \cot x+C\)

Let \( I=\int \frac{\sin ^6 x+\cos ^6 x}{\sin ^2 x \cos ^2 x} d x\)
\(\Rightarrow \quad I=\int \frac{\left(\sin ^2 x\right)^3+\left(\cos ^2 x\right)^3}{\sin ^2 x \cos ^2 x} d x\)
\(=\int \frac{\left[\begin{array}{l} \left(\sin ^2 x+\cos ^2 x\right)^3 \\ -3 \sin ^2 x \cos ^2 x\left(\sin ^2 x+\cos ^2 x\right) \end{array}\right]}{\sin ^2 x \cos ^2 x} d x\ \left[\because a^3+b^3=(a+b)^3-3 a b(a+b)\right]\)
\(=\int \frac{(1)^3-3 \sin ^2 x \cos ^2 x}{\sin ^2 x \cos ^2 x} d x\)
\(\left[\because \sin ^2 \theta+\cos ^2 \theta=1\right]\)
\(=\int \frac{1}{\sin ^2 x \cos ^2 x} d x-3 \int \frac{\sin ^2 x \cos ^2 x}{\sin ^2 x \cos ^2 x} d x\)
\(=\int \frac{\sin ^2 x+\cos ^2 x}{\sin ^2 x \cos ^2 x} d x-3 \int 1 d x\)
\(=\int\left[\frac{\sin ^2 x}{\sin ^2 x \cos ^2 x}+\frac{\cos ^2 x}{\sin ^2 x \cos ^2 x}\right] d x-3 \int 1 d x\)
\(=\int\left(\sec ^2 x+\operatorname{cosec}^2 x\right) d x-3 \int 1 d x\)
\(=\int \sec ^2 x d x+\int \operatorname{cosec}^2 x d x-3 \int 1 d x\)
\(=\tan x-\cot x-3 x+C\)

Let \(I=\int \frac{x+2}{\sqrt{x^2+5 x+6}} d x\)
Now, let us write, (x+2) as
\(x+2  =A \frac{d}{d x}\left(x^2+5 x+6\right)+B\)
\(\Rightarrow x+2 =A(2 x+5)+B\)
On equating the coefficients of x and constant terms from both sides, we get
\(2 A=1 \text { and } 5 A+B=2 \Rightarrow A=\frac{1}{2} \text { and } B=-\frac{1}{2}\)
\(\therefore \quad I  =\int \frac{\left\{\frac{1}{2}(2 x+5)-\frac{1}{2}\right\}}{\sqrt{x^2+5 x+6}} d x \)
\(=\frac{1}{2} \int \frac{2 x+5}{\sqrt{x^2+5 x+6}} d x-\frac{1}{2} \int \frac{1}{\sqrt{x^2+5 x+6}} d x \)
\(=\frac{1}{2} I_1-\frac{1}{2} I_2 \text { (say) }\)
Consider, \(I_1=\int \frac{2 x+5}{\sqrt{x^2+5 x+6}} d x\)
Put \(x^2+5 x+6=t \Rightarrow(2 x+5) d x=d t\)
\(\therefore I_1=\int \frac{1}{\sqrt{t}} d t=2 \sqrt{t}+C_1=2 \sqrt{x^2+5 x+6}+C_1\)
\(\left[\because t=x^2+5 x+6\right]\)
and \(I_2=\int \frac{1}{\sqrt{x^2+5 x+6}} d x\)
\(=\int \frac{1}{\sqrt{x^2+2 \times \frac{5}{2} \times x+6+\frac{25}{4}-\frac{25}{4}}} d x \)
\(=\int \frac{1}{\sqrt{\left(x+\frac{5}{2}\right)^2+6-\frac{25}{4}}} d x\)
\(= \int \frac{1}{\sqrt{\left(x+\frac{5}{2}\right)^2-\left(\frac{1}{2}\right)^2}} d x \)
\(= \log \left|\left(x+\frac{5}{2}\right)+\sqrt{\left(x+\frac{5}{2}\right)^2-\left(\frac{1}{2}\right)^2}\right|+C_2\) 
\({\left[\because \int \frac{d x}{\sqrt{x^2-a^2}}=\log \left|x+\sqrt{x^2-a^2}\right|+C\right] }\)
\(\Rightarrow \quad I_2=  \log \left|x+\frac{5}{2}+\sqrt{x^2+5 x+6}\right|+C_2 \quad \ldots \text { (iii) }\)
On putting the values of I₁ and I₂ from Eqs. (ii) and (iii) in Eq. (i), we get
\(I=  \frac{1}{2}\left[2 \sqrt{x^2+5 x+6}+C_1\right]  -\frac{1}{2}\left[\log \left|x+\frac{5}{2}+\sqrt{x^2+5 x+6}\right|+C_2\right]\)
\(=  \sqrt{x^2+5 x+6}+\frac{C_1}{2}  -\frac{1}{2} \log \left|x+\frac{5}{2}+\sqrt{x^2+5 x+6}\right|-\frac{C_2}{2} \)
\(\Rightarrow \quad I=  \sqrt{x^2+5 x+6}  -\frac{1}{2} \log \left|x+\frac{5}{2}+\sqrt{x^2+5 x+6}\right|+C\)
where \(C=\frac{C_1}{2}-\frac{C_2}{2}\).

Let \(I =\int e^x \frac{\left(x^2+1\right)}{(x+1)^2} d x\)
\( =\int e^x \frac{\left(x^2+1+2 x-2 x\right)}{(x+1)^2} d x\)
\(=\int e^x\left(\frac{(x+1)^2-2 x}{(x+1)^2}\right) d x \)
\( =\int e^x\left(1-\frac{2 x}{(x+1)^2}\right) d x\)
\( =\int e^x d x-2 \int e^x \cdot \frac{x}{(x+1)^2} d x\)
\(=e^x-2 \int e^x\left(\frac{x+1-1}{(x+1)^2}\right) d x\)
\(=e^x-2 \int e^x\left(\frac{1}{(x+1)}+\frac{(-1)}{(x+1)^2}\right) d x\)
Now, consider \(f(x)=\frac{1}{x+1}\), then \(f^{\prime}(x)=\frac{(-1)}{(x+1)^2}\)
Thus, the above integrand is of the form
\( e^x\left[f(x)+f^{\prime}(x)\right]\)
\(\therefore I=e^x-2 e^x \frac{1}{(x+1)}+C\)
\(\left[\because \int e^x\left[f(x)+f^{\prime}(x)\right] d x=e^x f(x)+C\right] \)
\(\Rightarrow \quad I=e^x\left(\frac{x+1-2}{x+1}\right)+C \Rightarrow \quad I=e^x\left(\frac{x-1}{x+1}\right)+C\)

Let \(I=\int e^{2 x} \sin (3 x+1) d x\)
\(=\sin (3 x+1) \int e^{2 x} d x-\int\left\{\frac{d}{d x} \sin (3 x+1) \int e^{2 x} d x\right\} d x\)
[by using integration by parts]
\(=\frac{\sin (3 x+1) \cdot e^{2 x}}{2}-\int 3 \cos (3 x+1) \cdot \frac{e^{2 x}}{2} d x\) 
\(=\frac{e^{2 x} \sin (3 x+1)}{2}-\frac{3}{2} \int e^{2 x} \cos (3 x+1) d x\)
\(=\frac{e^{2 x} \sin (3 x+1)}{2}-\frac{3}{2}\left[\cos (3 x+1) \int e^{2 x} d x\right.  \left.-\int\left\{\frac{d}{d x} \cos (3 x+1) \int e^{2 x} d x\right\} d x\right]\)
[again by using integration by parts]
\(=\frac{e^{2 x} \sin (3 x+1)}{2}-\frac{3}{2}\left[\cos (3 x+1) \cdot \frac{e^{2 x}}{2}\right. \left.-\int-3 \sin (3 x+1) \cdot \frac{e^{2 x}}{2} d x\right]+C_1\)
\(\Rightarrow I=\frac{e^{2 x} \sin (3 x+1)}{2}-\frac{3}{4} e^{2 x} \cos (3 x+1) -\frac{9}{4} \int e^{2 x} \sin (3 x+1) d x+C_1 \)
\(\Rightarrow I=\frac{e^{2 x} \sin (3 x+1)}{2}-\frac{3}{4} e^{2 x} \cos (3 x+1)-\frac{9}{4} I+C_1\)
[from Eq. (i)]
\(\Rightarrow \frac{13}{4} I=\frac{e^{2 x} \sin (3 x+1)}{2}-\frac{3 e^{2 x} \cos (3 x+1)}{4}+C_1\)
\(\therefore \quad I=\frac{2 e^{2 x} \sin (3 x+1)}{13}-\frac{3 e^{2 x} \cos (3 x+1)}{13}+C\)
where \(C=\frac{4 C_1}{13}\).