$I=\int \frac{2 x-1}{\sqrt{4 x-x^2}} d x$

Solution:

$2 x-1=A \frac{d}{d x}\left(4 x-x^2\right)+B$

$2 x-1=A(4-2 x)+B$

$2=-2 A \Rightarrow A=-1,$

$4 A+B=-1 \Rightarrow B=3$

$I=\int \frac{-1(4-2 x)+3}{\sqrt{4 x-x^2}} d x$

$=-\int \frac{4-2 x}{\sqrt{4 x-x^2}} d x+3 \int \frac{d x}{\sqrt{4 x-x^2}}$

[$4 x-x^2=-\left(x^2-4 x\right)=4-\left(x^2-4 x+4\right) =4-(x-2)^2$ ]

$I=\int \frac{x^2 d x}{\sqrt{x^6+a^6}} \quad a>0$

Solution:

$=\int \frac{x^2 d x}{\sqrt{\left(x^3\right)^2+a^6}}$

[$x^3=t \Rightarrow 3 x^2 d x=d t]$

$=\int \frac{1}{3} \frac{d t}{\sqrt{t^2+\left(a^3\right)^2}}$

$=\frac{1}{3}.\log \left|t+\sqrt{\left.t^2+a^3\right)^2}\right|+c$

$=\frac{1}{3} \log \left|x^3+\sqrt{x^6+a^6}\right|+c$

$I=\int \frac{P(x)}{Q(x)} d x$

$\int \frac{d x}{x^2-a^2};$

$P =1, Q=x^2-a^2$

$Q =(x-a)(x+a)$

Froms

1. $\frac{p x+q}{(x-a)(x-b)} \quad a \neq b$

2. $\frac{p x+q}{(x-a)^2}$

3. $\frac{p x^2+q x+r}{(x-a)(x-b)(x-c)},$

$a \neq b, b \neq c \ a \neq c$

4. $\frac{p x^2+q x+r}{\left.(x-a)^2(x-b\right)}$

5. $\frac{p x^2+a x+r}{(x-a_1)\left(a x^2+b x+c\right)}$

Partial fractions

1 $\frac{A}{x-a}+\frac{B}{x-b}$

$\int \frac{d x}{x^2-a^2}=\frac{1}{2 a} \log |\frac{x-a}{x+a}|+C$

Solution:

$\frac{1}{(x-a)(x+a)}=\frac{A}{(x-a)}+\frac{B}{(x+a)}$

$\Rightarrow 1=A(x+a)+B(x-a)$

$A+B=0, \quad a(A-B)=1$

$A=-B, \quad-2 a B=1$

$\Rightarrow B=-\frac{1}{2 a} \cdot A=\frac{1}{2 a}$

$\frac{1}{(x-a)(x+a)} =\frac{1}{2 a}\left[\frac{1}{x-a}-\frac{1}{x+a}\right]$

$=\frac{1}{2 a}[\log |x-a|-\log |x+a|]+c$

$\int \frac{d x}{x\left(x^4+1\right)}$

Solution

$= \int \frac{x^3 d x}{x^4\left(x^4+1\right)}$

[$x^4=t \Rightarrow 4 x^3 d x=d t \Rightarrow x^3 d x=\frac{d t}{4}$

$= \frac{1}{4} \int \frac{d t}{t(t+1)}$

$\{\frac{1}{t(t+1)}=\frac{A}{t}+\frac{B}{t+1} \Rightarrow 1=A(t+1)+B t \Rightarrow A=1, B=-1 \}$

$=\frac{1}{4} \int\left(\frac{1}{t}-\frac{1}{t+1}\right) d t$

$=\frac{1}{4}\left[{[\log |t|-\log |t+1|]} +C\right.$

$=\frac{1}{4}[{\log |\frac{x^4}{x^4 + 1}|}] +C.$

$I =\int \frac{\left(x^2+1\right)\left(x^2+2\right)}{\left(x^2+3\right)\left(x^2+4\right)}dx$

Solution:

Integrand $= \frac{(x^2+1)(x^2+2)}{(x^2+3)(x^2+4)}$

Put $x^2=y$

Integrand = $\frac{(y+1)(y+2)}{(y+3)(y+4)}=\frac{y^2+3 y+2}{y^2+7 y+12}$

$\frac{y^2+3 y+2}{y^2+7 y+12} =1+\frac{-4 y-10}{y^2+7 y+12}$

$\Rightarrow I=\int \frac{\left(x^2+1\right)\left(x^2+2\right)}{\left(x^2+3\right)\left(x^2+4\right)} dx$

$={\int\left[1-\frac{4 x^2+10}{\left(x^2+3\right)\left(x^2+4\right)}\right] d x}$

[$\frac{4 y+10}{(y+3)(y+4)}=\frac{A}{(y+3)}+\frac{B}{(y+4)}$

A=-2, B=6

$=\int 1 d x-\{\int \frac{-2 d x}{x^2+3}+\int \frac{6 d x}{x^2+4}\}$

$=x+2 \int \frac{1}{x^2+3} d x-6 \int \frac{1}{x^2+4} d x$

$=x + 2.\frac{1}{\sqrt{3}} \tan ^{-1} \frac{x}{\sqrt{3}} - 6.\frac{1}{2}.\tan ^{-1} \frac{x}{2}+C$

$I=\int \frac{x+1}{(x-2)\left(x^2+1\right)} d x$

$\frac{x+1}{(x-2)\left(x^2+1\right)}=\frac{A}{x-2}+\frac{B x+C}{x^2+1}$

$\Rightarrow x+1=A\left(x^2+1\right)+(B x+C)(x-2)$

$A+B=0,-2 B+C=1, A-2 C=1$

$A=\frac{3}{5}, B=-\frac{3}{5}, C=-1 / 5$

$I=\int\left[\frac{3}{5} \cdot \frac{1}{x-2}+\frac{-\frac{3}{5} x-\frac{1}{5}}{x^2+1}\right] d x$

$=\frac{3}{5} \int \frac{d x}{x-2}-\frac{1}{5} \int \frac{3 x d x}{x^2+1}-\frac{1}{5} \int \frac{1}{x^2+1} d x$

$=\frac{3}{5} \log |x-2|-\frac{3}{5} \times \frac{1}{2} \log \left|x^2+1\right|-\frac{1}{5} \tan ^{-1} x+C$

$I =\int x e^x d x$

$x$ - 1st function

$e^x$ - 2nd function

$I =x \cdot e^x-\int 1 \cdot e^x d x$

$I =x e^x-e^x+c$

Remark:-

$\int x e^x d x$

= $x \cdot\left(e^x+c\right)-\int 1 \cdot\left(e^x+c\right) d x$

= $x e^x+c x-e^x-c x+c_1$

= $x e^x-e^x+c_1$