Ex 7.5 Class 12 Maths Ncert Solutions

Let \( \frac{x}{(x+1)(x+2)}=\frac{A}{(x+1)}+\frac{B}{(x+2)} \)
\(\Rightarrow x=\mathrm{A}(x+2)+\mathrm{B}(x+1)\)
On comparing the coefficients of \( x \) and constant term, we get,
\(A+B=1\)
\(2 A+B=0\)
On solving above two equations, we get,
\(\mathrm{A}=-1 \text { and } \mathrm{B}=2\)
Thus,
\(\frac{x}{(x+1)(x+2)}=\frac{-1}{(x+1)}+\frac{2}{(x+2)}\)
\(=\int \frac{x}{(x+1)(x+2)}=\int\left\{\frac{-1}{(x+1)}+\frac{2}{(x+2)}\right\} d x\)
\(=-\log |x+1|+2 \log |x+2|+\mathrm{C}\)
\(=\log (x+2)^{2}-\log |x+1|+\mathrm{C}\)
\(=\log \frac{(x+2)^{2}}{(x+1)}+C\)

Now \( \frac{1}{x^{2}-9}=\frac{1}{(x+3)(x-3)} \)
Let \( \frac{1}{(x+3)(x-3)}=\frac{A}{(x+3)}+\frac{B}{(x-3)} \)
\(\Rightarrow 1=\mathrm{A}(x-3)+\mathrm{B}(x+3)\)
On comparing the coefficients of \( x \) and constant term, we get,
\(A+B=0\)
\(-3 A+3 B=1\)
On solving above two equations, we get,
\( A=-\frac{1}{6} \) and \( B=\frac{1}{6} \)
Thus,
\(\frac{1}{(x+3)(x-3)}=\frac{-1}{6(x+3)}+\frac{1}{6(x-3)}\)
\(\int \frac{1}{(x+3)(x-3)} d x=\int\left\{\frac{-1}{6(x+3)}+\frac{1}{6(x-3)}\right\} d x\)
\(=-\frac{1}{6} \log |x+3|+\frac{1}{6}|x-3|+C\)
\(=\frac{1}{6} \log \left|\frac{(x-3)}{(x+3)}\right|+C\)

Let \( \frac{3 x-1}{(x-1)(x-2)(x-3)}=\frac{A}{(x-1)}+\frac{B}{(x-2)}+\frac{C}{(x-3)} \)
\(\Rightarrow 3 x-1=\mathrm{A}(x-2)(x-3)+\mathrm{B}(x-1)(x-3)\)\(+\mathrm{C}(x-1)(x-2) \quad\ldots\)(1)
Substituting \( x=1,2 \) and 3 respectively in equation (1), we get, \( \mathrm{A}=1, \mathrm{~B}=-5 \) and \( \mathrm{C}=4 \)
Thus,
\(\frac{3 x-1}{(x-1)(x-2)(x-3)}=\frac{1}{(x-1)}-\frac{5}{(x-2)}+\frac{4}{(x-3)}\)
\(\int \frac{3 x-1}{(x-1)(x-2)(x-3)} d x=\int \frac{1}{(x-1)}-\frac{5}{(x-2)}+\frac{4}{(x-3)} d x\)
\(=\log |x-1|-5 \log |x-2|+4 \log |x-3|+C\)

Let \( \frac{x}{(x-1)(x-2)(x-3)}=\frac{A}{(x-1)}+\frac{B}{(x-2)}+\frac{C}{(x-3)} \)
\(\Rightarrow x=\mathrm{A}(x-2)(x-3)+\mathrm{B}(x-1)(x-3)\) \(+\mathrm{C}(x-1)(x-2) \quad\ldots \text{(1)}\)
Substituting \( x=1,2 \) and 3 respectively in equation (1), we get,
\( \mathrm{A}=\frac{1}{2}, \mathrm{~B}=-2 \) and \( \mathrm{C}=\frac{3}{2} \)
Thus,
\(\frac{x}{(x-1)(x-2)(x-3)}=\frac{1}{2(x-1)}-\frac{2}{(x-2)}+\frac{3}{2(x-3)}\)
\(\int \frac{x}{(x-1)(x-2)(x-3)} d x=\int\left\{\frac{1}{2(x-1)}-\frac{2}{(x-2)}+\frac{3}{2(x-3)}\right\} d x\)
\(=\frac{1}{2} \log |x-1|-2 \log |x-2|+\frac{3}{2} \log |x-3|+\mathrm{C}\)

Let \( \frac{2 x}{x^{2}+3 x+2}=\frac{A}{(x+1)}+\frac{B}{(x+2)} \)
\(\Rightarrow 2 x=\mathrm{A}(x+2)+\mathrm{B}(x+1) \quad\ldots \text{(1)}\)
Substituting \( x=-1 \) and \(-2\) respectively in equation (1), we get,
\(A=-2, B=4\)
Thus, \( \frac{2 x}{x^{2}+3 x+2}=\frac{-2}{(x+1)}+\frac{4}{(x+2)} \)
\(\int \frac{2 x}{x^{2}+3 x+2} d x=\int\left\{\frac{-2}{(x+1)}+\frac{4}{(x+2)}\right\} d x\)
\(=4 \log |x+2|-2 \log |x+1|+\mathrm{C}\)

\(\frac{1-x^{2}}{x(1-2 x)}\)
On dividing \( 1-x^{2} \) by \( x(1-2 x) \), we get,
\(\frac{1-x^{2}}{x(1-2 x)}=\frac{1}{2}+\frac{1}{2}\left(\frac{2-x}{x(1-2 x)}\right)\quad\ldots \text{(1)}\)
Now, let \( \left(\frac{2-x}{x(1-2 x)}\right)=\frac{A}{x}+\frac{B}{(1-2 x)} \)
\((2-x)=A(1-2 x)+B x\quad\ldots \text{(2)}\)
Now, substituting \( x=0 \) and \( \frac{1}{2} \) in equation (2), we get,
\( \mathrm{A}=2 \) and \( \mathrm{B}=3 \)
Thus, \( \frac{2-x}{x(1-2 x)}=\frac{2}{x}+\frac{3}{(1-2 x)} \)
Now, putting this value in equation (2), we get,
\(\int \frac{1-x^{2}}{x(1-2 x)} d x=\int\left\{\frac{1}{2}+\frac{1}{2}\left(\frac{2}{x}+\frac{3}{(1-2 x)}\right)\right\} d x\)
\(=\frac{x}{2}+\log |x|+\frac{3}{2(-2)} \log |1-2 x|+C\)
\(=\frac{x}{2}+\log |x|-\frac{3}{4} \log |1-2 x|+C\)

Let \( \frac{x}{\left(x^{2}+1\right)(x-1)}=\frac{A x+B}{\left(x^{2}+1\right)}+\frac{C}{(x-1)} \)
\(x=(\mathrm{A}x+\mathrm{B})(x-1)+\mathrm{C}\left(x^{2}+1\right)\)
\(\Rightarrow x=\mathrm{Ax}^{2}-\mathrm{A}x+\mathrm{B}x-\mathrm{B}+\mathrm{C}x^{2}+\mathrm{C}\)
Equating the coefficients of \( x^{2}, x \) and constant term, we get,
\(A+C=0\)
\(-A+B=0\)
\(-B+C=0\)
On solving these equation, we get,
\( \mathrm{A}=-\frac{1}{2}, \mathrm{~B}=\frac{1}{2} \) and \( \mathrm{C}=\frac{1}{2} \)
Thus,
\(\frac{x}{\left(x^{2}+1\right)(x-1)}=\frac{-\frac{1}{2} x+\frac{1}{2}}{\left(x^{2}+1\right)}+\frac{\frac{1}{2}}{(x-1)}\)
\(\int \frac{x}{\left(x^{2}+1\right)(x-1)} d x=-\frac{1}{2} \int \frac{x}{x^{2}+1} d x+\frac{1}{2} \int \frac{1}{x^{2}+1} d x\)\(+\frac{1}{2} \int \frac{1}{x-1} d x\)
\(=-\frac{1}{4} \int \frac{2 x}{x^{2}+1} d x+\frac{1}{2} \tan ^{-1} x+\frac{1}{2} \log |x-1|+C\)
Now, let us consider, \( \int \frac{2 x}{x^{2}+1} d x \),
Let \( \left(x^{2}+1\right)=t \)
\(2 x\mathrm{d}x=\mathrm{dt}\)
Thus,
\(\int \frac{2 x}{x^{2}+1} d x=\int \frac{d t}{t}=\log |t|=\log \left|x^{2}+1\right|\)
Therefore,
\(\int \frac{x}{\left(x^{2}+1\right)(x-1)} d x=-\frac{1}{4} \log \left|x^{2}+1\right|+\frac{1}{2} \tan ^{-1} x\)\(+\frac{1}{2} \log |x-1|+C\)
\(=\frac{1}{2} \log |x-1|-\frac{1}{4} \log \left|x^{2}+1\right|+\frac{1}{2} \tan ^{-1} x+C\)

Let \( \frac{x}{(x-1)^{2}(x+2)}=\frac{A}{(x-1)}+\frac{B}{(x-1)^{2}}+\frac{C}{(x+2)} \)
\(\Rightarrow x=\mathrm{A}(x-1)(x+2)+\mathrm{B}(x+2)+\)\(\mathrm{C}(x-1)^{2} \quad\ldots \text{(1)}\)
Substituting \( x=-1 \) in equation (1), we get, \( \mathrm{B}=\frac{1}{3} \)
Equating the coefficients of \( x^{2} \) and constant term, we get,
\(\mathrm{A}+\mathrm{C}=0\)
\(-2 \mathrm{A}+2 \mathrm{B}+\mathrm{C}=0\)
\(\mathrm{A}=\frac{3}{9} \text { and } \mathrm{c}=\frac{-2}{9}\)
Thus,
\(\frac{x}{(x-1)^{2}(x+2)}=\frac{2}{9(x-1)}+\frac{1}{3(x-1)^{2}}+\frac{2}{9(x+2)}\)
\(\int \frac{x}{(x-1)^{2}(x+2)} d x=\int \frac{2}{9(x-1)}+\frac{1}{3(x-1)^{2}}+\frac{2}{9(x+2)} d x\)
\(=\frac{2}{9} \int \frac{1}{(x-1)} d x+\frac{1}{3} \int \frac{1}{(x-1)^{2}} d x-\frac{2}{9} \int \frac{1}{(x+2)} d x\)
\(=\frac{2}{9} \log |x-1|+\frac{1}{3}\left(\frac{-1}{x-1}\right)-\frac{2}{9} \log |x+2|+C\)
\(=\frac{2}{9} \log \left|\frac{x-1}{x+2}\right|+\frac{1}{3}\left(\frac{-1}{x-1}\right)+C\)

\(\frac{3 x+5}{x^{3}-x^{2}-x+1}=\frac{3 x+5}{(x-1)^{2}(x+1)}\)
Let \( \frac{3 x+5}{x^{3}-x^{2}-x+1}=\frac{A}{(x-1)}+\frac{B}{(x-1)^{2}}+\frac{C}{(x+1)} \)
\(\Rightarrow 3 x+5=\mathrm{A}(x-1)(x+1)+\mathrm{B}(x+1)+\mathrm{C}(x-1)^{2}\)
\(\Rightarrow 3 x+5=\mathrm{A}\left(x^{2}-1\right)+\mathrm{B}(x+1)+\mathrm{C}\)\(\left(x^{2}+1-2 x\right) \quad\ldots \text{(1)}\)
Substituting \( x=1 \) in equation (1), we get,
\(B=4\)
Equating the coefficients of \(x^{2}\) and \(x\), we get,
\( \mathrm{A}+\mathrm{C}=0 \)
\( b-2 \mathrm{C}=3 \)
\( \mathrm{A}=-\frac{1}{2} \) and \( \mathrm{c}=\frac{1}{2} \)
Thus,
\(\frac{3 x+5}{x^{3}-x^{2}-x+1}=\frac{-1}{2(x-1)}+\frac{4}{(x-1)^{2}}+\frac{1}{2(x+1)}\)
\(\int \frac{3 x+5}{x^{3}-x^{2}-x+1} d x=\int\left\{\frac{-1}{2(x-1)}+\frac{4}{(x-1)^{2}}+\frac{1}{2(x+1)}\right\} d x\)
\(=\frac{-1}{2} \int \frac{1}{(x-1)} d x+4 \int \frac{1}{(x-1)^{2}} d x+\frac{1}{2} \int \frac{1}{(x+1)}+C\)
\(=\frac{1}{2} \log \left|\frac{x+1}{x-1}\right|-\frac{4}{x-1}+C\)

\(\frac{2 x-3}{\left(x^{2}-1\right)(2 x+3)}=\frac{2 x-3}{(x+1)(x-1)(2 x+3)}\)
Let \( \frac{2 x-3}{(x+1)(x-1)(2 x+3)}=\frac{A}{(x+1)}+\frac{B}{(x-1)}+\frac{C}{(2 x+3)} \)
\(\Rightarrow 2 x-3=\mathrm{A}(x-1)(2 x+3)+\mathrm{B}(x+1)(2 x+3)+\)\(\mathrm{C}(x+1)(x-1)\)
\(\Rightarrow 2 x-3=\mathrm{A}\left(2 x^{2}+x-3\right)+\mathrm{B}\left(2 x^{2}+5 x+3\right)+\)\(\mathrm{C}\left(x^{2}-1\right)\)
\(\Rightarrow 2 x-3=(2 \mathrm{A}+2 \mathrm{B}+\mathrm{C}) x^{2}+(\mathrm{A}+5 \mathrm{~B}) x+\)\((-3 \mathrm{A}+3 \mathrm{B}-\mathrm{C})\)
Equating the coefficients of \(x^{2}\) and \(x\), we get,
\( \mathrm{B}=-\frac{1}{10}, \mathrm{~A}=\frac{5}{2} \) and \( \mathrm{C}=-\frac{24}{5} \)
Thus,
\(\frac{2 x-3}{(x+1)(x-1)(2 x+3)}=\frac{5}{2(x+1)}-\frac{1}{10(x-1)}-\frac{24}{5(2 x+3)}\)
\(\int \frac{2 x-3}{(x+1)(x-1)(2 x+3)} d x=\int \frac{5}{2(x+1)}-\frac{1}{10(x-1)}-\frac{24}{5(2 x+3)} d x\)
\(=\frac{5}{2} \int \frac{1}{(x+1)} d x-\frac{1}{10} \int \frac{1}{(x-1)} d x-\frac{24}{5} \int \frac{1}{(2 x+3)} d x\)
\(=\frac{5}{2} \log |x+1|-\frac{1}{10} \log |x-1|-\frac{24}{5 \times 2} \log |2 x+3|+C\)
\(=\frac{5}{2} \log |x+1|-\frac{1}{10} \log |x-1|-\frac{24}{10} \log |2 x+3|+C\)

\(\frac{5 x}{(x+1)\left(x^{2}-4\right)}=\frac{5 x}{(x+1)(x+2)(x-2)}\)
Let \( \frac{5 x}{(x+1)(x+2)(x-2)}=\frac{A}{(x+1)}+\frac{B}{(x+2)}+\frac{C}{(x-2)} \)
\(\Rightarrow 5 x=\mathrm{A}(x+2)(x-2)+\mathrm{B}(x+1)(x-2)+\)\(\mathrm{C}(x+1)(x+2)\quad\ldots \text{(1)}\)
Substituting \( x=-1,-2 \) and 2 respectively in equation (1), we get, \( \mathrm{A}=\frac{5}{3}, \mathrm{~B}=-\frac{5}{2} \), and \( \mathrm{C}=\frac{5}{6} \)
Thus,
\(\frac{5 x}{(x+1)(x+2)(x-2)}=\frac{5}{3(x+1)}-\frac{5}{2(x+2)}+\frac{5}{6(x-2)}\)
\(\int \frac{5 x}{(x+1)(x+2)(x-2)} d x=\int\left\{\frac{5}{3(x+1)}-\frac{5}{2(x+2)}+\frac{5}{6(x-2)}\right\} d x\)
\(=\frac{5}{3} \int \frac{1}{(x+1)} d x-\frac{5}{2} \int \frac{1}{(x+2)} d x+\frac{5}{6} \int \frac{1}{(x-2)} d x\)
\(=\frac{5}{3} \log |x+1|-\frac{5}{2} \log |x+2|+\frac{5}{6} \log |x-2|+C\)

\(\frac{x^{3}+x+1}{x^{2}-1}\)
Dividing \( \left(x^{3}+x+1\right) \) by \( x^{2}-1 \), we get,
\( \frac{x^{3}+x+1}{x^{2}-1}=x+\frac{2 x+1}{x^{2}-1} \)
Let \( \frac{2 x+1}{x^{2}-1}=\frac{A}{(x+1)}+\frac{B}{(x-1)} \)
Now, \( 2 x+1=\mathrm{A}(x-1)+\mathrm{B}(x+1) \quad\ldots \text{(1)} \)
Substituting \( x=1 \) and \(-1\) in equation (1), we get, \( \mathrm{A}=\frac{1}{2} \) and \( \mathrm{B}=\frac{3}{2} \)
Thus, \( \frac{x^{3}+x+1}{x^{2}-1}=x+\frac{1}{2(x+1)}+\frac{3}{2(x-1)} \)
\(\int \frac{x^{3}+x+1}{x^{2}-1}=\int x d x+\frac{1}{2} \int \frac{1}{(x+1)} d x+\frac{3}{2} \int \frac{1}{(x-1)} d x\)
\(=\frac{x^{2}}{2}+\frac{1}{2} \log |x+1|+\frac{3}{2} \log |x-1|+C\)

Let \( \frac{2}{(1-x)\left(1+x^{2}\right)}=\frac{A}{(1-x)}+\frac{B}{\left(1+x^{2}\right)} \)
\(\Rightarrow 2=\mathrm{A}\left(1-x^{2}\right)+(\mathrm{B}x+\mathrm{C})(1-x)\)
\(\Rightarrow 2=\mathrm{A}+\mathrm{A}x^{2}+\mathrm{B}x-\mathrm{B}x^{2}+\mathrm{C}-\mathrm{Cx}\)
On comparing the coefficients of \( x^{2}, x \) and constant term, we get,
\( \mathrm{A}-\mathrm{B}=0 \)
\( \mathrm{B}-\mathrm{C}=0 \)
\( \mathrm{A}+\mathrm{C}=0 \)
On solving these equations, we get,
\(\mathrm{A}=1, \mathrm{~B}=1 \text { and } \mathrm{C}=1\)
Thus,
\(\frac{2}{(1-x)\left(1+x^{2}\right)}=\frac{1}{(1-x)}+\frac{x+1}{\left(1+x^{2}\right)}\)
\(\int \frac{2}{(1-x)\left(1+x^{2}\right)} d x=\int \frac{1}{(1-x)} d x+\int \frac{x}{\left(1+x^{2}\right)} d x+\int \frac{1}{\left(1+x^{2}\right)} d x\)
\(-\log |x-1|+\frac{1}{2} \log \left|1+x^{2}\right|+\tan ^{-1} x+C\)

Let \( \frac{3 x-1}{(x+2)^{2}}=\frac{A}{(x+2)}+\frac{B}{(x+2)^{2}} \)
\(\Rightarrow 3 x-1=\mathrm{A}(x+2)+\mathrm{B}\)
Equating the coefficients of \(x\) and constant term, we get,
\(\mathrm{A}=3\)
\(2 \mathrm{A}+\mathrm{B}=-1\)
\(B=-7\)
Thus,
\(\frac{3 x-1}{(x+2)^{2}}=\frac{3}{(x+2)}-\frac{7}{(x+2)^{2}}\)
\(\int \frac{3 x-1}{(x+2)^{2}}=\int\left\{\frac{3}{(x+2)}-\frac{7}{(x+2)^{2}}\right\} d x\)
\(=3 \log |x+2|-7\left(\frac{-1}{(x+2)}\right)+C\)
\(=3 \log |x+2|+\frac{7}{(x+2)}+C\)

\(\frac{1}{x^{4}-1}=\frac{1}{\left(x^{2}-1\right)\left(x^{2}+1\right)}=\frac{1}{(x+1)(x-1)\left(x^{2}+1\right)}\)
Let \( \frac{1}{(x+1)(x-1)\left(x^{2}+1\right)}=\frac{A}{(x+1)}+\frac{B}{(x-1)}+\frac{c x+D}{\left(x^{2}+1\right)} \)
\(1=\mathrm{A}(x-1)\left(x^{2}+1\right)+\mathrm{B}(x+1)\left(x^{2}+1\right)+\)\((\mathrm{C}x+\mathrm{D})\left(x^{2}-1\right)\)
\(1=\mathrm{A}\left(x^{3}+x-x^{2}-1\right)+\mathrm{B}\left(x^{3}+x+x^{2}+1\right)\)\(+\mathrm{C}x^{3}+\mathrm{D}x^{2}-\mathrm{C}x-\mathrm{D}\)
\(1=(\mathrm{A}+\mathrm{B}+\mathrm{C}) x^{3}+(-\mathrm{A}+\mathrm{B}+\mathrm{D}) x^{2}+\)\((\mathrm{A}+\mathrm{B}-\mathrm{C}) x+(-\mathrm{A}+\mathrm{B}-\mathrm{D})\)
Equating the coefficients of \( x^{3}, x^{2}, x \) and constant term, we get,
\((A+B+C)=0\)
\((-A+B+C)=0\)
\((A+B-C)=0\)
\((-A+B-D)=0\)
On solving these equations, we get, \( \mathrm{A}=-\frac{1}{4}, B=\frac{1}{4}, C=0 \) and \( \mathrm{D}=-\frac{1}{2} \)
Therefore,
\(\frac{1}{x^{4}-1}=\frac{-1}{4(x+1)}+\frac{1}{4(x-1)}-\frac{1}{2\left(x^{2}+1\right)}\)
\(\int \frac{1}{x^{4}-1} d x=\int\left\{\frac{-1}{4(x+1)}+\frac{1}{4(x-1)}-\frac{1}{2\left(x^{2}+1\right)}\right\} d x\)
\(=-\frac{1}{4} \log |x-1|+\frac{1}{4} \log |x-1|-\frac{1}{2} \tan ^{-1} x+C\)
\(=\frac{1}{4} \log \left|\frac{x-1}{x+1}\right|-\frac{1}{2} \tan ^{-1} x+C\)

\(\frac{1}{\left(x^{n}+1\right)}\)
Multiplying numerator and denominator by \( x^{\mathrm{n}-1} \), we get,
\(\frac{1}{x\left(x^{n}+1\right)}=\frac{x^{n-1}}{x^{n-1} x\left(x^{n}+1\right)}=\frac{x^{n-1}}{x^{n}\left(x^{n}+1\right)}\)
Let \( x^{\mathrm{n}}=\mathrm{t} \)
\(x^{\mathrm{n}-1} \mathrm{d}x=\mathrm{dt}\)
Therefore,
\(\int \frac{1}{x\left(x^{n}+1\right)} d x=\int \frac{x^{n-1}}{x^{n-1} x\left(x^{n}+1\right)} d x=\frac{1}{n} \int \frac{1}{t(t+1)} d t\)
Let \( \frac{1}{t(t+1)}=\frac{A}{t}+\frac{B}{(t+1)} \)
\(1=\mathrm{A}(1+\mathrm{t})+\mathrm{Bt}\quad\ldots \text{(1)}\)
Substituting \( \mathrm{t}=0,-1 \) in equation (1), we get, \( \mathrm{A}=1 \) and \( \mathrm{B}=-1 \)
Thus,
\(\frac{1}{t(t+1)}=\frac{1}{t}+\frac{1}{(t+1)}\)
\(\int \frac{1}{x\left(x^{n}+1\right)} d x=\frac{1}{n} \int\left\{\frac{1}{t}+\frac{1}{(t+1)}\right\} d t\)
\(=\frac{1}{n}[\log |t|-\log |t+1|]+C\)
\(=\frac{1}{n}\left[\log \left|x^{n}\right|-\log \left|x^{n}+1\right|\right]+C\)
\(=\frac{1}{n} \log \left|\frac{x^{n}}{x^{n}+1}\right|+C\)

\(\frac{\cos x}{(1-\sin x)(2-\sin x)}\)
Let \( \sin x=\mathrm{t} \)
\(\cos x\mathrm{d}x=\mathrm{dt}\)
Therefore,
\(\int \frac{\cos x}{(1-\sin x)(2-\sin x)} d x=\int \frac{d t}{(1-t)(2-t)}\)
Let \( \int \frac{1}{(1-t)(2-t)}=\frac{A}{(1-t)}+\frac{B}{(2-t)} \)
\(1=A(2-t)+B(1-t) \ldots(1)\)
Substituting \( \mathrm{t}=2 \) and then \( \mathrm{t}=1 \) in equation (1), we get, Therefore,
\(\frac{1}{(1-t)(2-t)}=\frac{1}{(1-t)}+\frac{1}{(2-t)}\)
\(\int \frac{\cos x}{(1-\sin x)(2-\sin x)} d x=\int\left\{\frac{1}{(1-t)}+\frac{1}{(2-t)}\right\} d t\)
\(=-\log |1-t|+\log |2-t|+C\)
\(=\log \left|\frac{2-t}{1-t}\right|+C\)
\(=\log \left|\frac{2-\sin x}{1-\sin x}\right|+C\)

Now, \( \frac{\left(x^{2}+1\right)\left(x^{2}+2\right)}{\left(x^{2}+3\right)\left(x^{2}+4\right)}=1-\frac{\left(4 x^{2}+10\right)}{\left(x^{2}+3\right)\left(x^{2}+4\right)} \)
Let \( \frac{\left(4 x^{2}+10\right)}{\left(x^{2}+3\right)\left(x^{2}+4\right)}=\frac{A x+B}{\left(x^{2}+3\right)}+\frac{C x+D}{\left(x^{2}+4\right)} \)
\(4 x^{2}+10=(A x+B)\left(x^{2}+4\right)+(C x+D)\left(x^{2}+3\right)\)
\(\Rightarrow 4 x^{2}+10=A x^{3}+4 A x+B x^{2}+4 B+\)\(C x^{3}+3 C x+D x^{2}+3 D\)
\(\Rightarrow 4 x^{2}+10=(A+C) x^{3}+(B+D) x^{2}+\)\((4 A+3 C) x+(4 B+3 D)\)
Equating the coefficients of \( x^{3}, x^{2}, x \) and constant term, we get,
\(A+C=0\)
\(B+D=0\)
\(4 A+3 C=0\)
\(4 B+3 B=0\)
On solving these equations, we get, \( \mathrm{A}=0, \mathrm{~B}=-2, \mathrm{C}=0 \) and \( \mathrm{D}=6 \)
Therefore,
\(\frac{\left(4 x^{2}+10\right)}{\left(x^{2}+3\right)\left(x^{2}+4\right)}=\frac{-2}{\left(x^{2}+3\right)}+\frac{6}{\left(x^{2}+4\right)}\)
\(\frac{\left(x^{2}+1\right)\left(x^{2}+2\right)}{\left(x^{2}+3\right)\left(x^{2}+4\right)}=1-\left(\frac{-2}{\left(x^{2}+3\right)}+\frac{6}{\left(x^{2}+4\right)}\right)\)
\(\int \frac{\left(x^{2}+1\right)\left(x^{2}+2\right)}{\left(x^{2}+3\right)\left(x^{2}+4\right)} d x=\int\left\{1+\frac{-2}{\left(x^{2}+3\right)}-\frac{6}{\left(x^{2}+4\right)}\right\} d x\)
\(=\int\left\{1+\frac{2}{x^{2}+(\sqrt{3})^{2}}-\frac{6}{\left(x^{2}+2^{2}\right)}\right\}\)
\(=x+2\left(\frac{1}{\sqrt{3}} \tan ^{-1} \frac{x}{\sqrt{3}}\right)-6\left(\frac{1}{2} \tan ^{-1} \frac{x}{2}\right)+C\)
\(=x+\frac{2}{\sqrt{3}} \tan ^{-1} \frac{x}{\sqrt{3}}-3 \tan ^{-1} \frac{x}{2}+C\)

Now, \( \frac{2 x}{\left(x^{2}+1\right)\left(x^{2}+3\right)} \)
Now, let \( x^{2}=\mathrm{t} \)
\(2x \mathrm{d}x=\mathrm{dt}\)
Thus, \( \int \frac{2 x}{\left(x^{2}+1\right)\left(x^{2}+3\right)} d x=\int \frac{d t}{(t+1)(t+3)} \quad\ldots \text{(1)} \)
Let \( \frac{1}{(t+1)(t+3)}=\frac{A}{t+1}+\frac{B}{t+3} \)
\(1=\mathrm{A}(\mathrm{t}+3)+\mathrm{B}(\mathrm{t}+1)\quad\ldots \text{(2)}\)
Substituting \( \mathrm{t}=-3 \) and \(-1\) in (2), we get \( A=\frac{1}{2} \) and \( B=-\frac{1}{2} \)
Therefore, \( \frac{1}{(t+1)(t+3)}=\frac{1}{2(t+1)}+\frac{1}{2(t+3)} \)
\(\int \frac{2 x}{\left(x^{2}+1\right)\left(x^{2}+3\right)} d x=\int\left\{\frac{1}{2(t+1)}+\frac{1}{2(t+3)}\right\} d t\)
\(=\frac{1}{2} \log |(t+1)|-\frac{1}{2} \log |t+3|+C\)
\(=\frac{1}{2} \log \left|\frac{t+1}{t+3}\right|+C\)
\(=\frac{1}{2} \log \left|\frac{x^{2}+1}{x^{2}+3}\right|+C\)

Now, \( \frac{1}{x\left(x^{4}-1\right)} \)
Multiplying numerator and denominator by \( x^{3} \), we get,
\(\frac{1}{x\left(x^{4}-1\right)}=\frac{x^{3}}{x^{4}\left(x^{4}-1\right)}\)
Therefore, \( \int \frac{1}{x\left(x^{4}-1\right)} d x=\int \frac{x^{3}}{x^{4}\left(x^{4}-1\right)} d x \)
Now, let \( x^{4}=\mathrm{t} \)
\(4 x^{3} d x=d t\)
Thus, \( \int \frac{1}{x\left(x^{4}-1\right)} d x=\frac{1}{4} \int \frac{d t}{t(t-1)} \)
Let \( \frac{1}{t(t-1)}=\frac{A}{t}+\frac{B}{t-1} \)
\(1=\mathrm{A}(\mathrm{t}-1)+\mathrm{Bt} \quad\ldots \text{(1)}\)
Substituting \( \mathrm{t}=0 \) and 1 in (1), we get
\( A=-1 \) and \( B=1 \)
Therefore, \( \frac{1}{t(t-1)}=\frac{-1}{t}+\frac{1}{t-1} \)
\(\int \frac{1}{x\left(x^{4}-1\right)} d x=\frac{1}{4} \int\left\{\frac{-1}{t}+\frac{1}{t-1}\right\} d t\)
\(=\frac{1}{4}[\log |t|+\log |t-1|]+C\)
\(=\frac{1}{4} \log \left|\frac{t-1}{t}\right|+C\)
\(=\frac{1}{4} \log \left|\frac{x^{4}-1}{x^{4}}\right|+C\)

Now, \( \frac{1}{\left(e^{x}-1\right)} \)
Let \( \mathrm{e}^{x}=\mathrm{t} \)
\( e^{x} d x=d t \)
\( \int \frac{1}{\left(e^{x}-1\right)} d x=\int \frac{1}{t-1} \times \frac{d t}{t}=\int \frac{1}{t(t-1)} d t \)
Let \( \frac{1}{t(t-1)}=\frac{A}{t}+\frac{B}{t-1} \)
\( 1=\mathrm{A}(\mathrm{t}-1)+\mathrm{Bt} \ldots(1) \)
Substituting \( \mathrm{t}=1 \) and \( \mathrm{t}=0 \) in equation (1), we get
\( \mathrm{A}=-1 \) and \( \mathrm{B}=1 \)
Therefore, \( \frac{1}{t(t-1)}=\frac{-1}{t}+\frac{1}{t-1} \)
\( \int \frac{1}{t(t-1)} d t=\log \left|\frac{t-1}{t}\right|+C \)
\( =\log \left|\frac{e^{x}-1}{e^{x}}\right|+C \)

Let \( \frac{x}{(x-1)(x-2)}=\frac{A}{x-1}+\frac{B}{x-2} \)
\( x=\mathrm{A}(x-2)+\mathrm{B}(x-1) \quad\ldots \text{(1)} \)
Substituting \( x=1 \) and 2 in (1), we get,
\( \mathrm{A}=-1 \) and \( \mathrm{B}=2 \)
Therefore, \( \frac{x}{(x-1)(x-2)}=-\frac{1}{(x-1)}+\frac{2}{(x-2)} \)
\( \int \frac{x}{(x-1)(x-2)} d x=\int\left\{-\frac{1}{(x-1)}+\frac{2}{(x-2)}\right\} d x \)
\(=-\log |x-1|+2 \log |x-2|+\mathrm{C}\)
\(=\log \left|\frac{(x-2)^{2}}{x-1}\right|+C\)

Let \( \frac{1}{x\left(x^{2}+1\right)}=\frac{A}{x}+\frac{B x+C}{x^{2}+1} \)
\(1=A\left(x^{2}+1\right)+(B x+C)\)
Equating the coefficients of \( x^{2}, x \) and constant term, we get,
\(A+B=0\)
\(C=0\)
\(A=1\)
On solving these equations, we get,
\( \mathrm{A}=1, \mathrm{~B}=-1 \) and \( \mathrm{C}=0 \)
Therefore, \( \frac{1}{x\left(x^{2}+1\right)}=\frac{1}{x}+\frac{-x}{x^{2}+1} \)
\(\int \frac{1}{x\left(x^{2}+1\right)}=\int \frac{1}{x}+\frac{-x}{x^{2}+1} d x\)
\(=\log |x|-\frac{1}{2} \log \left|x^{2}+1\right|+C\)