Let \(I=\int \frac{3 x+5}{x^2+3 x-18} d x\)
Also, let \(3 x+5=A \frac{d}{d x}\left(x^2+3 x-18\right)+B\)
\(\Rightarrow \quad 3 x+5=A(2 x+3)+B\)
On comparing the coefficient of x, we get
\(2 A=3 \Rightarrow A=\frac{3}{2}\)and on comparing the constant terms, we get
\(B=5-3 A \Rightarrow B=5-3\left(\frac{3}{2}\right)=\frac{1}{2}\)
From Eq. (ii), we get
\(3 x+5=\frac{3}{2}(2 x+3)+\frac{1}{2}\)
From Eqs. (i) and (iii), we get
\(I =\int \frac{\frac{3}{2}(2 x+3)+\frac{1}{2}}{x^2+3 x-18} d x\)
\(=\frac{3}{2} \int \frac{2 x+3}{x^2+3 x-18} d x+\frac{1}{2} \int \frac{1}{x^2+3 x-18} d x \)
\(=\frac{3}{2} \log \left|x^2+3 x-18\right|+\frac{1}{2} \int \frac{1}{\left(x+\frac{3}{2}\right)^2-\frac{81}{4}} d x\)
\(=\frac{3}{2} \log \left|x^2+3 x-18\right|+\frac{1}{2} \int \frac{1}{\left(x+\frac{3}{2}\right)^2-\left(\frac{9}{2}\right)^2} d x\)
\(=\frac{3}{2} \log \left|x^2+3 x-18\right|+\frac{1}{2} \cdot \frac{1}{2\left(\frac{9}{2}\right)} \log \left|\frac{\left(x+\frac{3}{2}\right)-\frac{9}{2}}{\left(x+\frac{3}{2}\right)+\frac{9}{2}}\right|+C\)
\({\left[\because \int \frac{d x}{x^2-a^2}=\frac{1}{2 a} \log \left|\frac{x-a}{x+a}\right|+C\right]}\)
\(=\frac{3}{2} \log \left|x^2+3 x-18\right|+\frac{1}{18} \log \left|\frac{x-3}{x+6}\right|+C\)

We will solve this integral by partial fraction.
\(\text { Let } \frac{x^2}{\left(x^2+1\right)\left(3 x^2+4\right)}=\frac{A x+B}{\left(x^2+1\right)}+\frac{C x+D}{\left(3 x^2+4\right)} \)
\( \Rightarrow x^2=(A x+B)\left(3 x^2+4\right)+(C x+D)\left(x^2+1\right) \)
\(=(3 A+C) x^3+(3 B+D) x^2+(4 A+C) x+4 B+D\)
Equating similar terms, we get
3 A+C=0,3 B+D=1,4 A+C=0 and 4 B+D=0
On solving, we get A=0, B =-1, C=0 and D=4
Thus, \(I  =\int \frac{-d x}{\left(x^2+1\right)}+\int \frac{4 d x}{\left(3 x^2+4\right)}\)
\(=\int-\frac{d x}{\left(x^2+1\right)}+\frac{4}{3} \int \frac{d x}{\left(x^2+\frac{4}{3}\right)}\)
\(=\int-\frac{d x}{x^2+1}+\frac{4}{3} \int \frac{d x}{x^2+\left(\frac{2}{\sqrt{3}}\right)^2}\)
\(=-\tan ^{-1} x+\frac{4}{3} \cdot \frac{\sqrt{3}}{2} \tan ^{-1}\left(\frac{\sqrt{3} x}{2}\right)+C\)

Let \(I=\int_{0}^{\pi / 2}(2 \log |\sin x|-\log |\sin 2 x|) d x\) 
\( =\int_{0}^{\pi / 2}\left[\log \left(\sin ^{2} x\right)-\log (\sin 2 x)\right.] d x \\ \left[\because m \log n=\log n^{m}\right] \) 
\(=\int_{0}^{\pi / 2} \log \left(\frac{\sin ^{2} x}{\sin 2 x}\right) d x\left[\because \log m-\log n=\log \frac{m}{n}\right]\) 
\(=\int_{0}^{\pi / 2} \log \left(\frac{\sin ^{2} x}{2 \sin x \cos x}\right) d x\left[\because \sin ^{2} x=2 \sin x \cos x\right]\) 
\(=\int_{0}^{\pi / 2} \log \left(\frac{\tan x}{2}\right) d x=\int_{0}^{\pi / 2}[\log (\tan x)-\log 2] d x\) 
\(=\int_{0}^{\pi / 2} \log (\tan x) d x-\int_{0}^{\pi / 2} \log 2 d x\) 
\(=\int_{0}^{\pi / 2} \log (\tan x) d x-\log 2 \int_{0}^{\pi / 2} 1 d x\) 
\(\Rightarrow I=I_{1}-(\log 2)[x]_{0}^{\pi / 2}=I_{1}-\left(\frac{\pi}{2}-0\right) \log 2\)   ...(ii)
where, \(I_{1}=\int_{0}^{\pi / 2} \log (\tan x) d x\) 
Now \(I_{1}=\int_{0}^{\pi / 2} \log \left[\tan \left(\frac{\pi}{2}-x\right)\right] d x\) \(\left[\because \int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x\right]\) 
\(\Rightarrow I_{1}=\int_{0}^{\pi / 2} \log (\cot x) d x\)  ...(iii) 
On adding Eqs. (ii) and (iii), we get
\(2 I_{1}=\int_{0}^{\pi / 2}[\log (\tan x)+\log (\cot x)] d x\) 
\( =\int_{0}^{\pi / 2} \log (\tan x \cot x) d x \ [\because \log m+\log n=\log (m n)] \) 
\(=\int_{0}^{\pi / 2} \log 1 d x=0 \quad\left[\because \tan x=\frac{1}{\cot x}\right]\) 
\(\Rightarrow I_{1}=0\) 
From Eq. (i), we get \(I=0-\frac{\pi}{2} \log 2\) 
\(\Rightarrow I=-\frac{\pi}{2} \log 2\) 

Let \(I=\int_{-\pi}^{\pi}(\cos a x-\sin b x)^{2} d x\) 
\(=2 \int_{0}^{\pi}(\cos a x-\sin b x)^{2} d x\) 
\(\left[\begin{array}{c} \because \int_{-a}^{a} f(x) d x=\left\{\begin{array}{c} 2 \int_{0}^{a} f(x) d x, \text { if } f(x) \text { is even } \\ 0, & \text { if } f(x) \text { is odd } \end{array}\right. \\ \text { and the given integrand is an even function } \end{array}\right]\) 
\(=2 \int_{0}^{\pi}\left[\cos ^{2} a x+\sin ^{2} b x-2 \sin b x \cos a x\right] d x\) 
\( =2 \int_{0}^{\pi}\left[\cos ^{2} a x+\sin ^{2} b x-\{\sin (b x+a x)\right. +\sin (b x-a x)\}] d \) 
\([\because 2 \sin A \cos B=\sin (A+B)+\sin (A-B)]\) 
\(=2\left[\begin{array}{rl} \int_{0}^{\pi} \cos ^{2} a x d x+\int_{0}^{\pi} \sin ^{2} b x d x & -\int_{0}^{\pi} \sin (b+a) x d x \\ & -\int_{0}^{\pi} \sin (b-a) x d x \end{array}\right]\) 
\( =2\left[\int_{0}^{\pi}\left(\frac{1+\cos 2 a x}{2}\right) d x+\int_{0}^{\pi}\left(\frac{1-\cos 2 b x}{2}\right) d x\right. \left.-\int_{0}^{\pi} \sin (b+a) x d x-\int_{0}^{\pi} \sin (b-a) x d x\right] \) 
\( =2\left[\int_{0}^{\pi}\left(\frac{1}{2}+\frac{\cos 2 a x}{2}\right) d x+\int_{0}^{\pi}\left(\frac{1}{2}-\frac{\cos 2 b x}{2}\right) d x\right. \left.-\int_{0}^{\pi} \sin (b+a) x d x-\int_{0}^{\pi} \sin (b-a) x d x\right] \) 
\( =2\left[\int_{0}^{\pi}\right.\left(\frac{1}{2}+\frac{1}{2}\right) d x+\frac{1}{2} \int_{0}^{\pi} \cos 2 a x d x-\frac{1}{2} \int_{0}^{\pi} \cos 2 b x d x \left.-\int_{0}^{\pi} \sin (b+a) x d x-\int_{0}^{\pi} \sin (b-a) x d x\right] \) 
\( =2\left[[x]_{0}^{\pi}+\frac{1}{2}\right.\left[\frac{\sin 2 a x}{2 a}\right]_{0}^{\pi}-\frac{1}{2}\left[\frac{\sin 2 b x}{2 b}\right]_{0}^{\pi} +\left[\frac{\cos (b+a) x}{(b+a)}\right]_{0}^{\pi}+\left[\frac{\cos (b-a) x}{(b-a)}\right]_{0}^{\pi} \) 
\( = 2\left[\pi+\frac{1}{4 a}[\sin 2 a \pi]-\frac{1}{4 b}[\sin 2 b \pi]\right. \left.+\frac{1}{b+a} \cdot[\cos (b+a) \pi-1]+\frac{1}{(b-a)}[\cos (b-a) \pi-1]\right] \) 
\( =2\left[\pi+\frac{1}{4 a} \sin 2 a \pi-\frac{1}{4 b} \sin 2 b \pi+\frac{1}{b+a} \cos (b+a) \pi\right. \left.-\frac{1}{b+a}+\frac{1}{b-a} \cos (b-a) \pi-\frac{1}{b-a}\right] \)

Let \(I=\int_{0}^{1} \frac{\log (1+x)}{1+x^{2}} d x\)   ...(i)
Now, put \(x=\tan \theta \Rightarrow d x=\sec ^{2} \theta d \theta\) 
Lower limit When \(x=0, \text { then } \tan \theta=0 \Rightarrow \theta=0\) 
Upper limit When \(x=1, \text { then } \tan \theta=1 \Rightarrow \theta=\frac{\pi}{4}\) 
Now, \(I=\int_{0}^{\pi / 4} \frac{\log |1+\tan \theta|}{\left(1+\tan ^{2} \theta\right)} \sec ^{2} \theta d \theta\) 
\(=\int_{0}^{\pi / 4} \frac{\log |1+\tan \theta|}{\sec ^{2} \theta} \sec ^{2} \theta d \theta\left[\because \sec ^{2} \theta=1+\tan ^{2} \theta\right]\) 
\(I=\int_{0}^{\pi / 4} \log (1+\mid \tan \theta) d \theta\)   ...(ii)
\(=\frac{\pi}{8} \log 2\)

Let \(I=\int_{0}^{\pi} x \log |\sin x| d x\)   ...(i)  
\(\Rightarrow I=\int_{0}^{\pi}(\pi-x) \log |\sin (\pi-x)| d x\) 
\(=\int_{0}^{\pi}(\pi-x) \log |\sin x| d x\)   ..(ii)
On adding Eqs. (i) and (ii), we get
\(2 I=\pi \int_{0}^{\pi} \log |\sin x| d x\) 
\(\Rightarrow 2 I=2 \pi \int_{0}^{\pi / 2} \log |\sin x| d x\)    ...(iii)
\(\left[\because \int_{0}^{2 a} f(x) d x=2 \int_{0}^{a} f(x) d x, \text { if } f(2 a-x)=f(x)\right]\) 
\(\Rightarrow I=\pi \int_{0}^{\pi / 2} \log |\sin x| d x\)  ...(iv)
\(\Rightarrow I=\pi \int_{0}^{\pi / 2} \log |\sin (\pi / 2-x)| d x\) 
\(\left[\because \int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x\right]\) 
\(=\pi \int_{0}^{\pi / 2} \log |\cos x| d x\)   ...(v)
On adding Eqs. (iv) and (v), we get
\(2 I=\pi \int_{0}^{\pi / 2}(\log |\sin x|+\log |\cos x|) d x\) 
\(\Rightarrow 2 I I=\pi \int_{0}^{\pi / 2} \log |\sin x \cos x| d x\) 
\(\Rightarrow 2 I=\pi \int_{0}^{\pi / 2} \log \left|\frac{2 \sin x \cos x}{2}\right| d x\) 
[multiply by 2 from numerator and denominator]
\(\Rightarrow 2 I=\pi \int_{0}^{\pi / 2}(\log |\sin 2 x|-\log 2) d x\) 
\(2 I=\pi \int_{0}^{\pi / 2} \log |\sin 2 x| d x-\pi \int_{0}^{\pi / 2} \log 2 d x\) 
\(\Rightarrow 2 I=\pi \int_{0}^{\pi / 2} \log |\sin 2 x| d x-\pi \log 2[x]_{0}^{\pi / 2}\) 
Now, put \(2 x=t \Rightarrow d x=\frac{1}{2} d t\) 
Lower limit When \(x=0, \text { then } t=0\) 
Upper limit When \(x=\frac{\pi}{2}, \text { then } t=\pi\) 
\(\therefore \ 2 I=\frac{\pi}{2} \int_{0}^{\pi} \log |\sin t| d t-\frac{\pi^{2}}{2} \log 2\) 
\(\Rightarrow 2 I=\frac{\pi}{2} \int_{0}^{\pi} \log |\sin x| d x-\frac{\pi^{2}}{2} \log 2\) 
\(\Rightarrow 2 I=I-\frac{\pi^{2}}{2} \log 2\) [from ii]
\(\therefore I=-\frac{\pi^{2}}{2} \log 2=\frac{\pi^{2}}{2} \log \left(\frac{1}{2}\right)\)

Let \(I=\int_{0}^{\pi / 2} \frac{x \sin x \cos x}{\sin ^{4} x+\cos ^{4} x} d x\) 
Using \(\int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x\) we get
\(I=\int_{0}^{\pi / 2} \frac{\left(\frac{\pi}{2}-x\right) \sin \left(\frac{\pi}{2}-x\right) \cos \left(\frac{\pi}{2}-x\right)}{\sin ^{4}\left(\frac{\pi}{2}-x\right)+\cos ^{4}\left(\frac{\pi}{2}-x\right)} d x\) 
\(\Rightarrow I=\int_{0}^{\pi / 2} \frac{\left(\frac{\pi}{2}-x\right) \cos x \sin x}{\cos ^{4} x+\sin ^{4} x} d x\) 
\(\left[\because \cos \left(\frac{\pi}{2}-\theta\right)=\sin \theta \text { and } \sin \left(\frac{\pi}{2}-\theta\right)=\cos \theta\right]\) 
On adding Eqs. (i) and (ii), we get
\(2 I=\frac{\pi}{2} \int_{0}^{\pi / 2} \frac{\cos x \sin x}{\sin ^{4} x+\cos ^{4} x} d x\) 
\(\Rightarrow I=\frac{\pi}{4} \int_{0}^{\pi / 2} \frac{\sin x \cos x}{\left(\sin ^{2} x\right)^{2}+\left(1-\sin ^{2} x\right)^{2}} d x\) \(\left[\because \cos ^{2} \theta=1-\sin ^{2} \theta\right]\) 
Now, put \(\sin ^{2} x=t \Rightarrow 2 \sin x \cos x d x=d t\) 
\(\Rightarrow \sin x \cos x d x=\frac{d t}{2}\) 
Lower limit When \(x=0, \text { then } t=\sin ^{2} 0=0\) 
Upper limit When \(x=\frac{\pi}{2}, \text { then } t=\sin ^{2} \frac{\pi}{2}=1\) 
Now, \(I=\frac{\pi}{4} \int_{0}^{1} \frac{1}{t^{2}+(1-t)^{2}} \frac{d t}{2}\) 
\(\Rightarrow I=\frac{\pi}{8} \int_{0}^{1} \frac{1}{t^{2}+\left(1+t^{2}-2 t\right)} d t\) 
\(\Rightarrow I=\frac{\pi}{8} \int_{0}^{1} \frac{1}{2 t^{2}-2 t+1} d t \Rightarrow I=\frac{\pi}{16} \int_{0}^{1} \frac{1}{t^{2}-t+\frac{1}{2}} d\) 
\(\Rightarrow I=\frac{\pi}{16} \int_{0}^{1} \frac{1}{t^{2}-t+\left(\frac{1}{2}\right)^{2}-\left(\frac{1}{2}\right)^{2}+\frac{1}{2}} d t\) 

Let \(I=\int_{-1}^{1} \frac{x^{3}+|x|+1}{x^{2}+2|x|+1} d x\) 
\(=\int_{-1}^{1} \frac{x}{x^{2}+2|x|+1} d x+\int_{-1}^{1} \frac{|x|+1}{x^{2}+2|x|+1} d x\) 
\(=0+2 \int_{0}^{1} \frac{|x|+1}{x^{2}+2|x|+1} d x\) 
\(\left[\because \int_{-1}^{1} f(x) d x=\left\{\begin{array}{c} 2 \int_{0}^{1} f(x) d x, \text { if } f(x) \text { is even function } \\ 0, \quad \text { if } f(x) \text { is odd function } \end{array}\right]\right.\) 
\(=2 \int_{0}^{1} \frac{(x+1)}{x^{2}+2 x+1} d x \quad[\because|x|=x, \text { if } x \geq 0]\) 
\(=2 \int_{0}^{1} \frac{(x+1)}{(x+1)^{2}} d x\) 
= 2log2

(i) Put cos x = t.
(ii) Use the integral of the form \(\int \frac{(p x+q)}{a x^{2}+b x+c} d x\) 
= \(\log \left(\frac{9}{8}\right)\)

Let \(I=\int_{0}^{\pi / 2} \sin 2 x \tan ^{-1}(\sin x) d x\) 
\(=\int_{0}^{\pi / 2} 2 \sin x \cos x \tan ^{-1}(\sin x) d x\) \([\because \sin 2 x=2 \sin x \cos x]\) 
Now, put \(x=t \Rightarrow \cos x d x=d t\) 
Lower limit When \(x=0, \text { then } t=0\) 
Upper Iimit When \(x=\frac{\pi}{2}, \text { then } t=\sin \frac{\pi}{2}=1\) 
Now,\(I=\int_{0}^{1} 2 t \tan ^{-1} t d t=2 \int_{0}^{1} t \cdot \tan ^{-1} t d t\) 
\(=2\left(\left[\left(\tan ^{-1} t\right) \frac{t^{2}}{2}\right]_{0}^{1}-\int_{0}^{1} \frac{1}{1+t^{2}} \cdot \frac{t^{2}}{2} d t\right)\) 
[using integration by parts]
\(=2\left[\frac{\tan ^{-1} 1}{2}-0\right]-\int_{0}^{1} \frac{t^{2}}{1+t^{2}} d t\) 
\(=2\left(\frac{\pi}{8}\right)-\int_{0}^{1} \frac{\left(1+t^{2}\right)-1}{1+t^{2}} d t\) 
[adding and subtracting from numerator] 
\(=\frac{\pi}{4}-\int_{0}^{1}\left(1-\frac{1}{1+t^{2}}\right) d t=\frac{\pi}{4}-\left[t-\tan ^{-1} t\right]_{0}^{1}\) 
\(=\frac{\pi}{4}-\left[1-\tan ^{-1} 1-(0-0)\right]\) 
\(=\frac{\pi}{4}-1+\frac{\pi}{4}=\frac{\pi}{2}-1\)