Again ${{x}^{2}}=t$ $\frac{2t+1}{t(t+4)}=\frac{A}{t}+\frac{B}{(t+4)}.......(1)$ $2t+1=A(t+4)+B(t)$ Putting $t=-4$ $2(-4)+1=A(-4+4)+B(-4)$ $-8+1=0-4B$ $-7=-4B$ $B=\frac{7}{4}$ Putting $t=0$...

Put $\frac{2}{(1-x)(1+{{x}^{2}})}=\frac{A}{1-x}+\frac{Bx+C}{{{x}^{2}}+1}.....(1)$ $A\left( 1+{{x}^{2}} \right)+Bx(1-x)+C(1-x)=2$ Put $x=1$ $2=2A+0+0$ $A=1$ Put $x=0$ $2=A+C$ $C=2-A$ $C=2-1=1$...

Put $t=cos2x$ $dt=-2sin2xdx$ $I=\int \frac{-dt/2}{\left( \left( 1-t \right)\left( 2-t \right) \right)}=\frac{1}{2}\int \frac{dt}{(t-2)(1-t)}$ Putting...

Putting $\frac{{{\left( {{x}^{2}} \right)}^{2}}}{\left( {{x}^{2}}+1 \right)\left( {{x}^{2}}+9 \right)\left( {{x}^{2}}+16 \right)}=\frac{{{t}^{2}}}{(t+1)(t+9)(t+16)}$...

Putting $\frac{{{x}^{2}}}{\left( {{x}^{4}}-{{x}^{2}}-12 \right)}=\frac{t}{{{t}^{2}}-t-12}=\frac{t}{(t-4)(t+3)}$ $=\frac{A}{t-4}+\frac{B}{t+3}.....(1)$ Where $t={{x}^{2}}$ $A(t+3)+B(t-4)=t$ Now put...

Put $t=cosx$ $dt=-sinxdx$ $\frac{-dt}{\sin x}=dx$ $I=\int \frac{-dt}{{{\sin }^{2}}x(1+2t)}$ $=\int \frac{dt}{(1-{{\cos }^{2}}x)(1+2t)}=\int \frac{dt}{(1-{{t}^{2}})(1+2t)}$ Putting,...

Put $t=sinx$ $dt=cosx$ $I=\int \frac{\sin x\cos x}{{{\cos }^{2}}x(1-\sin x)}dx$ $=\int \frac{tdt}{(1-{{\sin }^{2}}x)(1-t)}=\int \frac{tdt}{(1-{{t}^{2}})(1-t)}$ Putting...

$=\int \left( \frac{{{\sin }^{2}}x+{{\cos }^{2}}x}{\sin x{{\cos }^{2}}x} \right)dx$ $=\int \frac{{{\sin }^{2}}x}{\sin x{{\cos }^{2}}x}dx+\int \frac{{{\cos }^{2}}x}{\sin x{{\cos }^{2}}x}dx$ $=\int...

Put $t=sinx$ $dt=cosxdx$ $I=\int \frac{dt}{(1-{{\sin }^{2}}x)(5-4t)}=\int \frac{dt}{(1-{{t}^{2}})(5-4t)}$ $\frac{1}{(1-{{t}^{2}})(5-4t)}=\frac{1}{(1-t)(1+t)(5-4t)}$ Putting,...

Put $t=cosx$ $\frac{dt}{-\sin x}=dx$ $I=\int \frac{dt}{-\frac{\sin x}{\sin x(3+2t)}}$ $=-\int \frac{dt}{{{\sin }^{2}}x(3+2t)}=-\int \frac{dt}{(1-{{\cos }^{2}}x)(3+2t)}$ $=-\int...

Put $t={{x}^{6}}$ $dt=6{{x}^{5}}dx$ \[\int \frac{dt}{\frac{\left( 6{{x}^{5}} \right)}{x\left( t+1 \right)}}=\frac{1}{6}\int \frac{dt}{{{x}^{6}}(t+1)}=\frac{1}{6}\int \frac{dt}{t(t+1)}\]...

Put $t={{x}^{2}}$ $dt=5{{x}^{4}}dx$ Putting $\frac{1}{t\left( t+1 \right)}=\frac{A}{t}+\frac{B}{t+1}......(1)$ $A(t+1)+Bt=1$ Now put $t+1=0$ $t=-1$ $A(0)+B(-1)=1$ $B=-1$ Now put $t=0$...

${{e}^{x}}=t+1$ $dt={{e}^{x}}dx$ $\frac{dt}{{{e}^{x}}}=dx$ $\frac{dt}{t+1}=dx$ Put, $\frac{1}{(1-t){{t}^{2}}}=\frac{A}{t+1}+\frac{Bt+C}{{{t}^{2}}}$ $A\left( {{t}^{2}} \right)+(Bt+C)(t+1)=1$ put...

Putting $\frac{{{x}^{2}}+1}{({{x}^{2}}+4)({{x}^{2}}+25)}=\frac{t+1}{\left( t+4 \right)\left( t+25 \right)}$ $=\frac{A}{t+4}+\frac{B}{t+25}.....(1)$ Where $t={{x}^{2}}$ $(A+B)t+(25A+4B)=t+1$...

Put $\frac{1}{\left( {{x}^{2}}+2 \right)\left( {{x}^{2}}+4 \right)}=\frac{1}{(t+2)(t+4)}=\frac{A}{t+2}+\frac{B}{t+4}......(1)$ $A(t+4)+B(t+2)=1$ Put $t+4=0$ $t=-4$ $A(0)+B(-4+2)=1$ $B=-\frac{1}{2}$...

Put $\frac{17}{\left( 2x+1 \right)\left( {{x}^{2}}+4 \right)}=\frac{A}{2x+1}+\frac{Bx+C}{{{x}^{2}}+4}.......(1)$ $A\left( {{x}^{2}}+4 \right)+(Bx+C)(2x+1)=17$ Put $2x+1=0$ $x=-\frac{1}{2}$ $A\left(...

Put $\frac{1}{({{x}^{2}}+1){{(x+1)}^{2}}}=\frac{A}{x+1}+\frac{B}{{{(x+1)}^{2}}}+\frac{Cx+D}{{{x}^{2}}+1}.....(1)$ $A(x+1)({{x}^{2}}+1)+B({{x}^{2}}+1)+(Cx+D){{(x+1)}^{2}}=1$ Put $x+1=0$ $x=-1$...

Put $\frac{1}{{{x}^{3}}-1}=\frac{1}{(x+1)({{x}^{2}}-x+1)}=\frac{A}{x+1}+\frac{Bx+C}{{{x}^{2}}-x+1}.....(1)$ $A\left( {{x}^{2}}-x+1 \right)+(bx+C)(x+1)=1$ Now putting $x+1=0$ $x=-1$ $A(1+1+1)+C(0)=1$...

Put $\frac{1}{{{x}^{3}}-1}=\frac{1}{\left( x-1 \right)({{x}^{2}}+x+1)}=\frac{A}{x-1}+\frac{Bx+C}{{{x}^{2}}+x+1}.....(1)$ $A\left( {{x}^{2}}+x+1 \right)+(Bx+C)(x-1)=1$ Now putting, $x-1=0$ $x=1$...

Put $t={{x}^{2}}$ $dt=2xdx$ Now putting, $\frac{{{x}^{2}}}{\left( {{x}^{4}}-1 \right)}=\frac{t}{(t-1)(t+1)}=\frac{A}{t-1}+\frac{B}{t+1}......(1)$ $A(t+1)+B(t-1)=t$ Putting, $t+1=0$ $t=-1$...

Put $t={{x}^{2}}$ $dt=2xdx$ Now putting, $\frac{1}{(t+1)(t+3)}=\frac{A}{t+1}+\frac{B}{t+3}.......(1)$ $A(t+3)+B(t+1)=1$ Putting $t+3=0$ $x=-3$ $A(0)+B(-3+1)=1$ $B=-\frac{1}{2}$ Putting $t+1=0$...

Now putting, $\frac{3x+5}{\left( {{x}^{3}}-{{x}^{2}}+x-1 \right)}=\frac{A}{x-1}+\frac{Bx-C}{\left( {{x}^{2}}+1 \right)}......(1)$ $A\left( {{x}^{2}}+1 \right)+(Bx+C)(x-1)=3x+5$ Putting, $x-1=09$...

Now putting, $\frac{8}{(x+2)({{x}^{2}}+4)}=\frac{A}{x+2}+\frac{Bx+C}{({{x}^{2}}+4)}.....(1)$ $A\left( {{x}^{2}}+4 \right)+\left( Bx+C \right)(x+2)=8$ Putting, $x+2=0$ $x=-2$ $A(4+4)+0=8$ $A=1$ By...

Now putting, $\frac{5{{x}^{2}}18x+17}{{{(x-1)}^{2}}(2x-3)}=\frac{A}{(2x-3)}+\frac{B}{x-1}+\frac{C}{{{(x-1)}^{2}}}......(1)$ $A{{\left( x-1 \right)}^{2}}+B(2x-3)(x-1)+C(2x-3)=5{{x}^{2}}-18x+17$...

Now putting, $\frac{3x+1}{(x+2){{(x-2)}^{2}}}=\frac{A}{(x+2)}+\frac{B}{(x-2)}+\frac{C}{{{(x-2)}^{2}}}......(1)$ $A{{\left( x-2 \right)}^{2}}+B\left( x+2 \right)(x-2)+C(x+2)=3x+1$ Putting, $x-2=0$...

Now putting, $\frac{2x}{{{(2x+1)}^{2}}}=\frac{A}{(2x+1)}+\frac{B}{{{(2x+1)}^{2}}}....(1)$ $A(2x+1)+B=2x$ Putting $2x+1=0$ $x=\frac{-1}{2}$ $A(0)+B=-1$ By equating the coefficient of x, $2A=2$ $A=1$...

Now putting, $\frac{{{x}^{2}}+x+1}{(x+1)({{x}^{2}}+1)}=\frac{A}{(x+2)}+\frac{Bx+c}{({{x}^{2}}+1)}$ $A\left( {{x}^{2}}+1 \right)+(Bx+C)(x+2)={{x}^{2}}+x+1$...

Now putting, $\frac{{{x}^{2}}+1}{(x-3){{(x-1)}^{2}}}=\frac{A}{(x-3)}+\frac{B}{(x-1)}+\frac{C}{{{(x-1)}^{2}}}......(1)$ $A{{\left( x-1 \right)}^{2}}+B\left( x-3 \right)(x-1)+C\left( x-3...

Now putting, $\frac{{{x}^{2}}+1}{(x+3){{(x-1)}^{2}}}=\frac{A}{(x+3)}+\frac{B}{(x-1)}+\frac{C}{{{(x-1)}^{2}}}....(1)$ $A{{\left( x-1 \right)}^{2}}+B\left( x+3 \right)(x-1)+C(x+3)={{x}^{2}}+1$ Now put...

Now putting, $\frac{2x+9}{(x+2){{(x-3)}^{2}}}=\frac{A}{(x+2)}+\frac{B}{(x-3)}+\frac{C}{{{(x-3)}^{2}}}.......(1)$ $A{{\left( x-3 \right)}^{2}}+B(x+2)(x-3)+C(x+2)=2x+9$ Now put, $x-3=0$ Therefore,...

Now putting, $\frac{{{x}^{2}}+x+1}{(x+2){{(x+1)}^{2}}}=\frac{A}{(x+2)}+\frac{B}{(x+1)}+\frac{C}{{{(x+1)}^{2}}}.......(1)$ $A{{\left( x+1 \right)}^{2}}+B\left( x+2 \right)\left( x+1 \right)+C\left(...

\[=\int \frac{1}{2}dx+\int \frac{x}{x(2x-1)}dx=\frac{1}{x(2x-1)}dx\] \[I=\frac{1}{2}x+\frac{1}{2}\times \frac{\log \left| 2x-1 \right|}{2}-{{I}_{1}}......(1)\] Where ${{I}_{1}}=\int...

Putting $t={{x}^{4}}$ $dt=4{{x}^{3}}dx$ $I=\int \frac{{{x}^{3}}dx}{{{x}^{4}}({{x}^{4}}-1)}=\frac{1}{4}\times \int \frac{dt}{t\left( t-1 \right)}$ Now putting, $\frac{1}{t\left( t-1...

Putting $t={{e}^{x}}$ $dt={{e}^{x}}dx$ $I=\int \frac{dt}{t(t-1)}$ Now putting, $\frac{1}{t(t-1)}=\frac{A}{t}+\frac{B}{t-1}.....(1)$ $A\left( t-1 \right)+Bt=1$ Now put $t-1=0$ therefore, $t=1$...

Putting $t=sinx$ $dt=cosxdx$ $I=\int \frac{2t}{\left( 1+t \right)\left( 2+t \right)}dt$ Now putting, $\frac{2t}{(1+t)(2+t)}=\frac{A}{1+t}+\frac{B}{2+t}......(1)$ $A\left( 2+t \right)+B(1+t)=2t$ Now...

Putting $t=tanx$ $dt={{\sec }^{2}}xdx$ $I=\int \frac{dt}{({{t}^{3}}+4)}=\int \frac{dt}{t\left( {{t}^{2}}+4 \right)}$ Now, putting $\frac{1}{t\left( {{t}^{2}}+4...

Putting $t=cotx$ $\begin{align} & dt=-\cos e{{c}^{2}}xdx \\ & I=\int \frac{-dt}{(1-{{t}^{2}})}=-\int \frac{1}{(1-{{t}^{2}})}dt \\ \end{align}$ $=\frac{-1}{2}\log \left| \frac{1+\cot...

Putting $t=logx$ $dt=dx/x$ $I=\int \frac{2tdt}{\left( 2{{t}^{2}}-t-3 \right)}$ Now putting, $\frac{2t}{\left( 2{{t}^{2}}-t-3 \right)}=\frac{A}{2t-3}+\frac{B}{t+1}.....(1)$ $A(t+1)+B(2t-3)=2t$ Now...

Putting $t={{e}^{x}}$ $dt={{e}^{x}}dx$ $I=\int \frac{dt}{({{t}^{3}}-3{{t}^{2}}-t+3)}=,\int \frac{dt}{\left( {{t}^{2}}((t-3)(t-3)) \right)}$ $=\int \frac{dt}{({{t}^{2}}-1)(t-3)}$ Now putting,...

Putting $t={{e}^{x}}$ $dt={{e}^{x}}dx$ $I=\int \frac{dt}{({{t}^{2}}+5t+6)}$ Now putting, $\frac{1}{({{t}^{2}}+5t+6)}=\frac{A}{2+t}+\frac{B}{3+t}.....(1)$ $A\left( 3+t \right)+B(2+t)=1$ Now put...

Putting $t=cosx$ $dt=-sinxdx$ $I=\int \frac{\left( -dt \right)t}{{{t}^{2}}-t-2}=-\frac{tdt}{(t+1)(t-2)}$ Now putting , $\frac{-t}{(t+1)(t-2)}=\frac{A}{t+1}+\frac{B}{t-2}......(1)$ $A(t-2)+B(t+1)=-t$...

Putting $t=tanx$ $dt={{\sec }^{2}}xdx$ $I=\int \frac{dt}{(2+t)(3+y)}$, Now putting $\frac{1}{(3+t)(2+t)}=\frac{A}{2+t}+\frac{B}{3+t}.....(1)$ $A(3+t)+B(2+t)=1$ Now put $t+2=0$ Therefore, $t=-2$...

Putting $t=sinx$ $dt=\cos xdx$ $I=\int \frac{dt}{\left( 1+t \right)\left( 2+t \right)}$ Now putting, $\frac{1}{(1+t)(2+t)}=\frac{A}{1+t}+\frac{B}{2+t}.......(1)$ $A(2+t)+B(1+t)=1$ Now put, $t+1=0$...

$2xdx=dt$ $\int \frac{dt}{(1+t)(3+t)}=\frac{1}{2}\int \left( \frac{1}{1+t}-\frac{1}{3+t} \right)dt$ $\frac{1}{2}\left[ \log \left| 1+t \right|-\log \left| 3+t \right| \right]+c=\frac{1}{2}\log...

$\int \frac{2x+1}{\left( 1-x \right)\left( 4+x \right)}dx$ Putting $\frac{2x+1}{\left( 1-x \right)\left( 4+x \right)}dx=\frac{A}{1-x}+\frac{B}{4+x}......(1)$ $A\left( 4+x \right)+B(1-x)=2x+1$ Now...

$=\int \left( -x+\frac{-2}{1-{{x}^{2}}} \right)dx$ $=\int -xdx+\left( -2 \right)\int \frac{1}{1-{{x}^{2}}}dx$ $=\frac{-{{x}^{2}}}{2}-\log \left| \frac{1+x}{1-x} \right|+c$...

$=\int \left\{ \left( x+3 \right)+\frac{7x-6}{\left( x-1 \right)\left( x-2 \right)} \right\}dx$ $=\frac{{{x}^{2}}}{2}+3x+\int \frac{7x-6}{(x-1)(x-2)}dx$ $=\frac{{{x}^{2}}}{2}+3x+{{I}_{1}}......(1)$...

$=\int \left( -1+\frac{5x+1}{\left( x+2 \right)(x-1)} \right)dx$ $=\int -dx++\int \frac{5x+1}{(x+2)(x-1)}dx$ $=-x+{{I}_{1}}$ ${{I}_{1}}=\int \frac{5x+1}{\left( x+2 \right)\left( x-1 \right)}dx$ Put,...

$I=\int x+\frac{4x}{{{x}^{2}}-4}dx$ $I=\int xdx+\int \frac{4x}{{{x}^{2}}-4}dx$ $=\frac{{{x}^{2}}}{x}+\int \frac{4x}{(x-2)(x+2)}dx$ ${{I}_{1}}=\int \frac{4x}{(x-2)(x+2)}dx$...

$=I\int \left( 1+\frac{2}{{{x}^{2}}-1} \right)$ $I=\int dx+2\int \frac{1}{{{x}^{2}}-1}dx$ \[I=x+2\times \frac{1}{2}\log \left| \frac{x-1}{x+1} \right|+c\] \[I=x+\log \left| \frac{x-1}{x+1}...

Which implies, $I=\int dx+\int \frac{2x+1}{{{x}^{2}}+3x+2}dx$ Therefore, $I=x+{{I}_{1}}$ Where ${{I}_{1}}=\int \frac{2x+1}{{{x}^{2}}+3x+2}dx$ Putting $\frac{2x+1}{\left( x+1 \right)\left( x+2...

Putting $\frac{\left( 2x+5 \right)}{\left( x-2 \right)\left( x+1 \right)}=\frac{A}{x-2}+\frac{B}{x+1}...(1)$ Which implies $A\left( x+1 \right)+B(x-2)=2x+5$ Now put $x+1=0$ Therefore, $x=-1$...

Putting $\frac{2x-3}{(x-1)(x+1)(2x+3)}=\frac{A}{x-1}+\frac{B}{x+1}+\frac{C}{2x+3}.....(1)$ Which implies, $A(x+1)(2x+3)+B(x-1)(2x+3)+C(x-1)(x+1)=2x-3$ Now put $x+1=0$ Therefore, $x=-1$...

Putting, $\frac{(2x-1)}{(x-1)(x+2)(x-3)}=\frac{A}{x-1}+\frac{B}{x+2}+\frac{C}{x-3}....(1)$ Which implies, $A(x+2)(x-2)+B(x-1)(x-3)+C(x-1)(x+2)=2x-1$ Now put $x+2=0$ Therefore, $x=-2$...

Putting $\frac{1}{x(x-2)(x-4)}=\frac{A}{x}+\frac{B}{x-2}+\frac{C}{x-4}.....(1)$ Which implies, $A(x-2)(x-4)+Bx(x-4)+Cx(x-2)=1$ Now put $x-2=0$ Therefore, $x=2$ $A(0)+B\times 2(2-4)+C(0)=1$ $B\times...

Putting $\frac{x}{(x+2)(3-2x)}=\frac{A}{x+2}+\frac{B}{3-2x}....(1)$ Which implies $A(3-2x)+B(x+2)=x$ Now put, $3-2x=0$ Therefore, $x=3/2$ $A(0)+B(\frac{3}{2}+2)=\frac{3}{2}$ $B\left( \frac{7}{2}...

Suppose $\frac{2x+1}{(x+2)(x-3)}=\frac{A}{x+2}+\frac{B}{x-3}....(1)$ Which implies $2x=1=A(x-3)+B(x+2)$ Now put, $x-3=0$, $x=3$ $2\times 3+1=A(0)+B(3+2)$ So, $B=\frac{7}{5}$ Now put $x+2=0$, $x=-2$...

$\frac{1}{x(x+2)}=\frac{A}{x}+\frac{B}{x+2}....(1)$ Which implies $A(x+2)+Bx=1$, putting $x+2=0$ Therefore, $x=-2$, And $B=-0.5$ From equation (1) $\frac{1}{x(x+2)}=\frac{1}{2}\times...