Explanation:

(C) : Given,
\(\int \frac{2 x-1}{(x-1)(x+2)(x-3)} d x\)
Partial by \(v\) fraction method-
\(\frac{2 x-1}{(x-1)(x+2)(x-3)}=\frac{A}{x-1}+\frac{B}{x+2}+\frac{C}{x-3}\)
\(\frac{2 x-1}{(x-1)(x+2)(x-3)}\)
\(=\frac{\mathrm{A}\left(\mathrm{x}^{2}-3 \mathrm{x}+2 \mathrm{x}-6\right)+\mathrm{B}\left(\mathrm{x}^{2}-3 \mathrm{x}-1 \mathrm{x}+3\right)+\mathrm{C}\left(\mathrm{x}^{2}+2 \mathrm{x}-\mathrm{x}-2\right)}{(\mathrm{x}-1)(\mathrm{x}+2)(\mathrm{x}-3)}\)
\((2 x-1)=A\left(x^{2}-x-6\right)+B\left(x^{2}-4 x+3\right)+C\left(x^{2}+x-2\right)\)
\((2 \mathrm{x}-1)=(\mathrm{A}+\mathrm{B}+\mathrm{C}) \mathrm{x}^{2}+(-\mathrm{A}-4 \mathrm{~B}+\mathrm{C}) \mathrm{x}+(-6 \mathrm{~A}+3 \mathrm{~B}-2 \mathrm{C})\)
On comparing the terms, we get -
\(0=A+B+C 2=-A-4 B+C \tag{ii}\)
\(-1=-6 A+3 B-2 B C \tag{iii}\)
On solving equation (i), (ii), (iii)
\(A=-\frac{1}{6}, B=\frac{-1}{3} C=\frac{1}{2}\)
\(\int \frac{2 x-1}{(x-1)(x-2)(x-3)}=\int \frac{-1}{6(x-1)} d x+\int \frac{-1}{3(x+2)} d x+\int \frac{1}{2(x-3)} d x\) \(=-\frac{1}{6} \log (\mathrm{x}-1)-\frac{1}{3} \log (\mathrm{x}+2)+\frac{1}{2} \log (\mathrm{x}-3)+\mathrm{C}\)

Explanation:

(B) : Given,
\(\int(\sec x) \log (\sec x-\tan x) d x\)
Let, \(\mathrm{t}=\log (\sec \mathrm{x}-\tan \mathrm{x})\)
On differentiation both sides w.r.t. \(\mathrm{x}\), we get -
\(\frac{d t}{d x}=\frac{1}{(\sec x-\tan x)}\left(\sec x \tan x-\sec ^{2} x\right)\)
\(\frac{d t}{d x}=\frac{\sec x(\tan x-\sec x)}{(\sec x-\tan x)}\)
\(\mathrm{dt}=-\sec \mathrm{x} d \mathrm{x}\)
Then,
\(=\int-\mathrm{tdt}=\frac{-\mathrm{t}^{2}}{2}+\mathrm{C}\)
\(=-\frac{1}{2}\{\log (\operatorname{sex}-\tan \mathrm{x})\}^{2}+\mathrm{C}\)

Explanation:

(B) : Let,
\(I=\int \frac{2 x^{12}+5 x^{9}}{\left(x^{5}+x^{3}+1\right)^{3}} d x\)
\(I=\int \frac{2 x^{12}+5 x^{9}}{x^{15}\left(1+x^{-2}+x^{-5}\right)^{3}} d x\)
\(I=\int \frac{2 x^{-3}+5 x^{-6}}{\left(1+x^{-2}+x^{-5}\right)} d x\)
Let, \(\mathrm{t}=\left(1+\mathrm{x}^{-2}+\mathrm{x}^{-5}\right)\)
\(\frac{\mathrm{dt}}{\mathrm{dx}}=\left(-2 \mathrm{x}^{-3}-5 \mathrm{x}^{-6}\right)\)
\(\left(2 \mathrm{x}^{-3}+5 \mathrm{x}^{-6}\right) \mathrm{dx}=-\mathrm{dt}\)
\(\therefore \quad \mathrm{I}=\int \frac{\mathrm{dt}}{\mathrm{t}^{3}}\)
\(I=-\int \mathrm{t}^{-3} \mathrm{dt} \Rightarrow \mathrm{I}=\frac{-\mathrm{t}^{-3+1}}{-3+1}+\mathrm{C}\)
\(\mathrm{I}=\frac{-1}{2 \mathrm{t}^{2}}+\mathrm{C} \Rightarrow \mathrm{I}=\frac{\mathrm{x}^{10}}{2\left(\mathrm{x}^{5}+\mathrm{x}^{3}+1\right)^{2}}+\mathrm{C}\)

Explanation:

Explanation:

(B) : Given,
\(\int\left(1+x-\frac{1}{x}\right) e^{x+\frac{1}{x}} d x\)
\(=\int e^{\left(x+\frac{1}{x}\right)} d x+\int x\left(1-\frac{1}{x^{2}}\right) e^{\left(x+\frac{1}{x}\right)} d x\)
\(=\int e^{\left(x+\frac{1}{x}\right)} d x+x e^{x+\frac{1}{x}}-\int e^{x+\frac{1}{x}} d x\)
Using integration by parts
\(\left[\because \int\left(1-\frac{1}{\mathrm{x}^{2}}\right) \mathrm{e}^{\mathrm{x}+\frac{1}{\mathrm{x}}}=\mathrm{e}^{\mathrm{x}+\frac{1}{\mathrm{x}}}\right.\)
\(=x e^{x+\frac{1}{x}}+c\)

Explanation:

(C) : Given,
\(\int \frac{2 x-1}{(x-1)(x+2)(x-3)} d x\)
Partial by \(v\) fraction method-
\(\frac{2 x-1}{(x-1)(x+2)(x-3)}=\frac{A}{x-1}+\frac{B}{x+2}+\frac{C}{x-3}\)
\(\frac{2 x-1}{(x-1)(x+2)(x-3)}\)
\(=\frac{\mathrm{A}\left(\mathrm{x}^{2}-3 \mathrm{x}+2 \mathrm{x}-6\right)+\mathrm{B}\left(\mathrm{x}^{2}-3 \mathrm{x}-1 \mathrm{x}+3\right)+\mathrm{C}\left(\mathrm{x}^{2}+2 \mathrm{x}-\mathrm{x}-2\right)}{(\mathrm{x}-1)(\mathrm{x}+2)(\mathrm{x}-3)}\)
\((2 x-1)=A\left(x^{2}-x-6\right)+B\left(x^{2}-4 x+3\right)+C\left(x^{2}+x-2\right)\)
\((2 \mathrm{x}-1)=(\mathrm{A}+\mathrm{B}+\mathrm{C}) \mathrm{x}^{2}+(-\mathrm{A}-4 \mathrm{~B}+\mathrm{C}) \mathrm{x}+(-6 \mathrm{~A}+3 \mathrm{~B}-2 \mathrm{C})\)
On comparing the terms, we get -
\(0=A+B+C 2=-A-4 B+C \tag{ii}\)
\(-1=-6 A+3 B-2 B C \tag{iii}\)
On solving equation (i), (ii), (iii)
\(A=-\frac{1}{6}, B=\frac{-1}{3} C=\frac{1}{2}\)
\(\int \frac{2 x-1}{(x-1)(x-2)(x-3)}=\int \frac{-1}{6(x-1)} d x+\int \frac{-1}{3(x+2)} d x+\int \frac{1}{2(x-3)} d x\) \(=-\frac{1}{6} \log (\mathrm{x}-1)-\frac{1}{3} \log (\mathrm{x}+2)+\frac{1}{2} \log (\mathrm{x}-3)+\mathrm{C}\)

Explanation:

(B) : Given,
\(\int(\sec x) \log (\sec x-\tan x) d x\)
Let, \(\mathrm{t}=\log (\sec \mathrm{x}-\tan \mathrm{x})\)
On differentiation both sides w.r.t. \(\mathrm{x}\), we get -
\(\frac{d t}{d x}=\frac{1}{(\sec x-\tan x)}\left(\sec x \tan x-\sec ^{2} x\right)\)
\(\frac{d t}{d x}=\frac{\sec x(\tan x-\sec x)}{(\sec x-\tan x)}\)
\(\mathrm{dt}=-\sec \mathrm{x} d \mathrm{x}\)
Then,
\(=\int-\mathrm{tdt}=\frac{-\mathrm{t}^{2}}{2}+\mathrm{C}\)
\(=-\frac{1}{2}\{\log (\operatorname{sex}-\tan \mathrm{x})\}^{2}+\mathrm{C}\)

Explanation:

(B) : Let,
\(I=\int \frac{2 x^{12}+5 x^{9}}{\left(x^{5}+x^{3}+1\right)^{3}} d x\)
\(I=\int \frac{2 x^{12}+5 x^{9}}{x^{15}\left(1+x^{-2}+x^{-5}\right)^{3}} d x\)
\(I=\int \frac{2 x^{-3}+5 x^{-6}}{\left(1+x^{-2}+x^{-5}\right)} d x\)
Let, \(\mathrm{t}=\left(1+\mathrm{x}^{-2}+\mathrm{x}^{-5}\right)\)
\(\frac{\mathrm{dt}}{\mathrm{dx}}=\left(-2 \mathrm{x}^{-3}-5 \mathrm{x}^{-6}\right)\)
\(\left(2 \mathrm{x}^{-3}+5 \mathrm{x}^{-6}\right) \mathrm{dx}=-\mathrm{dt}\)
\(\therefore \quad \mathrm{I}=\int \frac{\mathrm{dt}}{\mathrm{t}^{3}}\)
\(I=-\int \mathrm{t}^{-3} \mathrm{dt} \Rightarrow \mathrm{I}=\frac{-\mathrm{t}^{-3+1}}{-3+1}+\mathrm{C}\)
\(\mathrm{I}=\frac{-1}{2 \mathrm{t}^{2}}+\mathrm{C} \Rightarrow \mathrm{I}=\frac{\mathrm{x}^{10}}{2\left(\mathrm{x}^{5}+\mathrm{x}^{3}+1\right)^{2}}+\mathrm{C}\)

Explanation:

Explanation:

(B) : Given,
\(\int\left(1+x-\frac{1}{x}\right) e^{x+\frac{1}{x}} d x\)
\(=\int e^{\left(x+\frac{1}{x}\right)} d x+\int x\left(1-\frac{1}{x^{2}}\right) e^{\left(x+\frac{1}{x}\right)} d x\)
\(=\int e^{\left(x+\frac{1}{x}\right)} d x+x e^{x+\frac{1}{x}}-\int e^{x+\frac{1}{x}} d x\)
Using integration by parts
\(\left[\because \int\left(1-\frac{1}{\mathrm{x}^{2}}\right) \mathrm{e}^{\mathrm{x}+\frac{1}{\mathrm{x}}}=\mathrm{e}^{\mathrm{x}+\frac{1}{\mathrm{x}}}\right.\)
\(=x e^{x+\frac{1}{x}}+c\)

Explanation:

(C) : Given,
\(\int \frac{2 x-1}{(x-1)(x+2)(x-3)} d x\)
Partial by \(v\) fraction method-
\(\frac{2 x-1}{(x-1)(x+2)(x-3)}=\frac{A}{x-1}+\frac{B}{x+2}+\frac{C}{x-3}\)
\(\frac{2 x-1}{(x-1)(x+2)(x-3)}\)
\(=\frac{\mathrm{A}\left(\mathrm{x}^{2}-3 \mathrm{x}+2 \mathrm{x}-6\right)+\mathrm{B}\left(\mathrm{x}^{2}-3 \mathrm{x}-1 \mathrm{x}+3\right)+\mathrm{C}\left(\mathrm{x}^{2}+2 \mathrm{x}-\mathrm{x}-2\right)}{(\mathrm{x}-1)(\mathrm{x}+2)(\mathrm{x}-3)}\)
\((2 x-1)=A\left(x^{2}-x-6\right)+B\left(x^{2}-4 x+3\right)+C\left(x^{2}+x-2\right)\)
\((2 \mathrm{x}-1)=(\mathrm{A}+\mathrm{B}+\mathrm{C}) \mathrm{x}^{2}+(-\mathrm{A}-4 \mathrm{~B}+\mathrm{C}) \mathrm{x}+(-6 \mathrm{~A}+3 \mathrm{~B}-2 \mathrm{C})\)
On comparing the terms, we get -
\(0=A+B+C 2=-A-4 B+C \tag{ii}\)
\(-1=-6 A+3 B-2 B C \tag{iii}\)
On solving equation (i), (ii), (iii)
\(A=-\frac{1}{6}, B=\frac{-1}{3} C=\frac{1}{2}\)
\(\int \frac{2 x-1}{(x-1)(x-2)(x-3)}=\int \frac{-1}{6(x-1)} d x+\int \frac{-1}{3(x+2)} d x+\int \frac{1}{2(x-3)} d x\) \(=-\frac{1}{6} \log (\mathrm{x}-1)-\frac{1}{3} \log (\mathrm{x}+2)+\frac{1}{2} \log (\mathrm{x}-3)+\mathrm{C}\)

Explanation:

(B) : Given,
\(\int(\sec x) \log (\sec x-\tan x) d x\)
Let, \(\mathrm{t}=\log (\sec \mathrm{x}-\tan \mathrm{x})\)
On differentiation both sides w.r.t. \(\mathrm{x}\), we get -
\(\frac{d t}{d x}=\frac{1}{(\sec x-\tan x)}\left(\sec x \tan x-\sec ^{2} x\right)\)
\(\frac{d t}{d x}=\frac{\sec x(\tan x-\sec x)}{(\sec x-\tan x)}\)
\(\mathrm{dt}=-\sec \mathrm{x} d \mathrm{x}\)
Then,
\(=\int-\mathrm{tdt}=\frac{-\mathrm{t}^{2}}{2}+\mathrm{C}\)
\(=-\frac{1}{2}\{\log (\operatorname{sex}-\tan \mathrm{x})\}^{2}+\mathrm{C}\)

Explanation:

(B) : Let,
\(I=\int \frac{2 x^{12}+5 x^{9}}{\left(x^{5}+x^{3}+1\right)^{3}} d x\)
\(I=\int \frac{2 x^{12}+5 x^{9}}{x^{15}\left(1+x^{-2}+x^{-5}\right)^{3}} d x\)
\(I=\int \frac{2 x^{-3}+5 x^{-6}}{\left(1+x^{-2}+x^{-5}\right)} d x\)
Let, \(\mathrm{t}=\left(1+\mathrm{x}^{-2}+\mathrm{x}^{-5}\right)\)
\(\frac{\mathrm{dt}}{\mathrm{dx}}=\left(-2 \mathrm{x}^{-3}-5 \mathrm{x}^{-6}\right)\)
\(\left(2 \mathrm{x}^{-3}+5 \mathrm{x}^{-6}\right) \mathrm{dx}=-\mathrm{dt}\)
\(\therefore \quad \mathrm{I}=\int \frac{\mathrm{dt}}{\mathrm{t}^{3}}\)
\(I=-\int \mathrm{t}^{-3} \mathrm{dt} \Rightarrow \mathrm{I}=\frac{-\mathrm{t}^{-3+1}}{-3+1}+\mathrm{C}\)
\(\mathrm{I}=\frac{-1}{2 \mathrm{t}^{2}}+\mathrm{C} \Rightarrow \mathrm{I}=\frac{\mathrm{x}^{10}}{2\left(\mathrm{x}^{5}+\mathrm{x}^{3}+1\right)^{2}}+\mathrm{C}\)

Explanation:

Explanation:

(B) : Given,
\(\int\left(1+x-\frac{1}{x}\right) e^{x+\frac{1}{x}} d x\)
\(=\int e^{\left(x+\frac{1}{x}\right)} d x+\int x\left(1-\frac{1}{x^{2}}\right) e^{\left(x+\frac{1}{x}\right)} d x\)
\(=\int e^{\left(x+\frac{1}{x}\right)} d x+x e^{x+\frac{1}{x}}-\int e^{x+\frac{1}{x}} d x\)
Using integration by parts
\(\left[\because \int\left(1-\frac{1}{\mathrm{x}^{2}}\right) \mathrm{e}^{\mathrm{x}+\frac{1}{\mathrm{x}}}=\mathrm{e}^{\mathrm{x}+\frac{1}{\mathrm{x}}}\right.\)
\(=x e^{x+\frac{1}{x}}+c\)

Explanation:

(C) : Given,
\(\int \frac{2 x-1}{(x-1)(x+2)(x-3)} d x\)
Partial by \(v\) fraction method-
\(\frac{2 x-1}{(x-1)(x+2)(x-3)}=\frac{A}{x-1}+\frac{B}{x+2}+\frac{C}{x-3}\)
\(\frac{2 x-1}{(x-1)(x+2)(x-3)}\)
\(=\frac{\mathrm{A}\left(\mathrm{x}^{2}-3 \mathrm{x}+2 \mathrm{x}-6\right)+\mathrm{B}\left(\mathrm{x}^{2}-3 \mathrm{x}-1 \mathrm{x}+3\right)+\mathrm{C}\left(\mathrm{x}^{2}+2 \mathrm{x}-\mathrm{x}-2\right)}{(\mathrm{x}-1)(\mathrm{x}+2)(\mathrm{x}-3)}\)
\((2 x-1)=A\left(x^{2}-x-6\right)+B\left(x^{2}-4 x+3\right)+C\left(x^{2}+x-2\right)\)
\((2 \mathrm{x}-1)=(\mathrm{A}+\mathrm{B}+\mathrm{C}) \mathrm{x}^{2}+(-\mathrm{A}-4 \mathrm{~B}+\mathrm{C}) \mathrm{x}+(-6 \mathrm{~A}+3 \mathrm{~B}-2 \mathrm{C})\)
On comparing the terms, we get -
\(0=A+B+C 2=-A-4 B+C \tag{ii}\)
\(-1=-6 A+3 B-2 B C \tag{iii}\)
On solving equation (i), (ii), (iii)
\(A=-\frac{1}{6}, B=\frac{-1}{3} C=\frac{1}{2}\)
\(\int \frac{2 x-1}{(x-1)(x-2)(x-3)}=\int \frac{-1}{6(x-1)} d x+\int \frac{-1}{3(x+2)} d x+\int \frac{1}{2(x-3)} d x\) \(=-\frac{1}{6} \log (\mathrm{x}-1)-\frac{1}{3} \log (\mathrm{x}+2)+\frac{1}{2} \log (\mathrm{x}-3)+\mathrm{C}\)

Explanation:

(B) : Given,
\(\int(\sec x) \log (\sec x-\tan x) d x\)
Let, \(\mathrm{t}=\log (\sec \mathrm{x}-\tan \mathrm{x})\)
On differentiation both sides w.r.t. \(\mathrm{x}\), we get -
\(\frac{d t}{d x}=\frac{1}{(\sec x-\tan x)}\left(\sec x \tan x-\sec ^{2} x\right)\)
\(\frac{d t}{d x}=\frac{\sec x(\tan x-\sec x)}{(\sec x-\tan x)}\)
\(\mathrm{dt}=-\sec \mathrm{x} d \mathrm{x}\)
Then,
\(=\int-\mathrm{tdt}=\frac{-\mathrm{t}^{2}}{2}+\mathrm{C}\)
\(=-\frac{1}{2}\{\log (\operatorname{sex}-\tan \mathrm{x})\}^{2}+\mathrm{C}\)

Explanation:

(B) : Let,
\(I=\int \frac{2 x^{12}+5 x^{9}}{\left(x^{5}+x^{3}+1\right)^{3}} d x\)
\(I=\int \frac{2 x^{12}+5 x^{9}}{x^{15}\left(1+x^{-2}+x^{-5}\right)^{3}} d x\)
\(I=\int \frac{2 x^{-3}+5 x^{-6}}{\left(1+x^{-2}+x^{-5}\right)} d x\)
Let, \(\mathrm{t}=\left(1+\mathrm{x}^{-2}+\mathrm{x}^{-5}\right)\)
\(\frac{\mathrm{dt}}{\mathrm{dx}}=\left(-2 \mathrm{x}^{-3}-5 \mathrm{x}^{-6}\right)\)
\(\left(2 \mathrm{x}^{-3}+5 \mathrm{x}^{-6}\right) \mathrm{dx}=-\mathrm{dt}\)
\(\therefore \quad \mathrm{I}=\int \frac{\mathrm{dt}}{\mathrm{t}^{3}}\)
\(I=-\int \mathrm{t}^{-3} \mathrm{dt} \Rightarrow \mathrm{I}=\frac{-\mathrm{t}^{-3+1}}{-3+1}+\mathrm{C}\)
\(\mathrm{I}=\frac{-1}{2 \mathrm{t}^{2}}+\mathrm{C} \Rightarrow \mathrm{I}=\frac{\mathrm{x}^{10}}{2\left(\mathrm{x}^{5}+\mathrm{x}^{3}+1\right)^{2}}+\mathrm{C}\)

Explanation:

Explanation:

(B) : Given,
\(\int\left(1+x-\frac{1}{x}\right) e^{x+\frac{1}{x}} d x\)
\(=\int e^{\left(x+\frac{1}{x}\right)} d x+\int x\left(1-\frac{1}{x^{2}}\right) e^{\left(x+\frac{1}{x}\right)} d x\)
\(=\int e^{\left(x+\frac{1}{x}\right)} d x+x e^{x+\frac{1}{x}}-\int e^{x+\frac{1}{x}} d x\)
Using integration by parts
\(\left[\because \int\left(1-\frac{1}{\mathrm{x}^{2}}\right) \mathrm{e}^{\mathrm{x}+\frac{1}{\mathrm{x}}}=\mathrm{e}^{\mathrm{x}+\frac{1}{\mathrm{x}}}\right.\)
\(=x e^{x+\frac{1}{x}}+c\)

Explanation:

(C) : Given,
\(\int \frac{2 x-1}{(x-1)(x+2)(x-3)} d x\)
Partial by \(v\) fraction method-
\(\frac{2 x-1}{(x-1)(x+2)(x-3)}=\frac{A}{x-1}+\frac{B}{x+2}+\frac{C}{x-3}\)
\(\frac{2 x-1}{(x-1)(x+2)(x-3)}\)
\(=\frac{\mathrm{A}\left(\mathrm{x}^{2}-3 \mathrm{x}+2 \mathrm{x}-6\right)+\mathrm{B}\left(\mathrm{x}^{2}-3 \mathrm{x}-1 \mathrm{x}+3\right)+\mathrm{C}\left(\mathrm{x}^{2}+2 \mathrm{x}-\mathrm{x}-2\right)}{(\mathrm{x}-1)(\mathrm{x}+2)(\mathrm{x}-3)}\)
\((2 x-1)=A\left(x^{2}-x-6\right)+B\left(x^{2}-4 x+3\right)+C\left(x^{2}+x-2\right)\)
\((2 \mathrm{x}-1)=(\mathrm{A}+\mathrm{B}+\mathrm{C}) \mathrm{x}^{2}+(-\mathrm{A}-4 \mathrm{~B}+\mathrm{C}) \mathrm{x}+(-6 \mathrm{~A}+3 \mathrm{~B}-2 \mathrm{C})\)
On comparing the terms, we get -
\(0=A+B+C 2=-A-4 B+C \tag{ii}\)
\(-1=-6 A+3 B-2 B C \tag{iii}\)
On solving equation (i), (ii), (iii)
\(A=-\frac{1}{6}, B=\frac{-1}{3} C=\frac{1}{2}\)
\(\int \frac{2 x-1}{(x-1)(x-2)(x-3)}=\int \frac{-1}{6(x-1)} d x+\int \frac{-1}{3(x+2)} d x+\int \frac{1}{2(x-3)} d x\) \(=-\frac{1}{6} \log (\mathrm{x}-1)-\frac{1}{3} \log (\mathrm{x}+2)+\frac{1}{2} \log (\mathrm{x}-3)+\mathrm{C}\)

Explanation:

(B) : Given,
\(\int(\sec x) \log (\sec x-\tan x) d x\)
Let, \(\mathrm{t}=\log (\sec \mathrm{x}-\tan \mathrm{x})\)
On differentiation both sides w.r.t. \(\mathrm{x}\), we get -
\(\frac{d t}{d x}=\frac{1}{(\sec x-\tan x)}\left(\sec x \tan x-\sec ^{2} x\right)\)
\(\frac{d t}{d x}=\frac{\sec x(\tan x-\sec x)}{(\sec x-\tan x)}\)
\(\mathrm{dt}=-\sec \mathrm{x} d \mathrm{x}\)
Then,
\(=\int-\mathrm{tdt}=\frac{-\mathrm{t}^{2}}{2}+\mathrm{C}\)
\(=-\frac{1}{2}\{\log (\operatorname{sex}-\tan \mathrm{x})\}^{2}+\mathrm{C}\)

Explanation:

(B) : Let,
\(I=\int \frac{2 x^{12}+5 x^{9}}{\left(x^{5}+x^{3}+1\right)^{3}} d x\)
\(I=\int \frac{2 x^{12}+5 x^{9}}{x^{15}\left(1+x^{-2}+x^{-5}\right)^{3}} d x\)
\(I=\int \frac{2 x^{-3}+5 x^{-6}}{\left(1+x^{-2}+x^{-5}\right)} d x\)
Let, \(\mathrm{t}=\left(1+\mathrm{x}^{-2}+\mathrm{x}^{-5}\right)\)
\(\frac{\mathrm{dt}}{\mathrm{dx}}=\left(-2 \mathrm{x}^{-3}-5 \mathrm{x}^{-6}\right)\)
\(\left(2 \mathrm{x}^{-3}+5 \mathrm{x}^{-6}\right) \mathrm{dx}=-\mathrm{dt}\)
\(\therefore \quad \mathrm{I}=\int \frac{\mathrm{dt}}{\mathrm{t}^{3}}\)
\(I=-\int \mathrm{t}^{-3} \mathrm{dt} \Rightarrow \mathrm{I}=\frac{-\mathrm{t}^{-3+1}}{-3+1}+\mathrm{C}\)
\(\mathrm{I}=\frac{-1}{2 \mathrm{t}^{2}}+\mathrm{C} \Rightarrow \mathrm{I}=\frac{\mathrm{x}^{10}}{2\left(\mathrm{x}^{5}+\mathrm{x}^{3}+1\right)^{2}}+\mathrm{C}\)

Explanation:

Explanation:

(B) : Given,
\(\int\left(1+x-\frac{1}{x}\right) e^{x+\frac{1}{x}} d x\)
\(=\int e^{\left(x+\frac{1}{x}\right)} d x+\int x\left(1-\frac{1}{x^{2}}\right) e^{\left(x+\frac{1}{x}\right)} d x\)
\(=\int e^{\left(x+\frac{1}{x}\right)} d x+x e^{x+\frac{1}{x}}-\int e^{x+\frac{1}{x}} d x\)
Using integration by parts
\(\left[\because \int\left(1-\frac{1}{\mathrm{x}^{2}}\right) \mathrm{e}^{\mathrm{x}+\frac{1}{\mathrm{x}}}=\mathrm{e}^{\mathrm{x}+\frac{1}{\mathrm{x}}}\right.\)
\(=x e^{x+\frac{1}{x}}+c\)