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

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

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

$A=x f\left.(x)\right|_{x=0}=x\left.\frac{2 x^{2}-x+4}{x\left(x^{2}+4\right)}\right|_{x=0}=\frac{4}{4}=+ 1$

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

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

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

$2 x^{2}-x+4=A x^{2}+4 A+B x^{2}+C x$

$2 x^{2}-x+4=1 x^{2}+4+B x^{2}+c x$

$2=1+B \Rightarrow B=1$

$-1=c \Rightarrow(c=-1)$

$f(x)=\frac{1}{x}+\frac{x-1}{x^{2}+4}$

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

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

$\begin{array}{l}{\int \frac{1}{x^{2}+4} d x} \\ {=\frac{1}{2} \tan ^{-1}\left(\frac{x}{2}\right)+C}\end{array}$

$\int \frac{x}{x^{2}+4} d x=\frac{1}{2} \int \frac{2 x}{u} d x=\frac{1}{2} \int \frac{d u}{u}=\frac{1}{2} \ln |u|=\frac{1}{2} \ln \left|x^{2}+4\right|+c_{2}$

let $u=x^{2}+4 \rightarrow d u=2 x d x$

$x = {-b \pm \sqrt{b^2-4ac} \over 2a}$

$=\ln |x|+c_{1}+\frac{1}{2} \tan ^{-1}\left(\frac{x}{2}\right)+c_{2}+\frac{1}{2} \ln \left|x^{2}+4\right|+c_{3}$

where $c_{1}+c_{2}+c_{3}=c$

$\int \frac{2 x^{2}-x+4}{x\left(x^{2}+4\right)}=\ln |x|+\frac{1}{2} \tan ^{-1}\left(\frac{x}{2}\right)+\frac{1}{2} \ln \left|x^{2}+4\right|+c$

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

$\begin{array}{r}{4 x ^ { 2 } - 4 x + 3 | { 4 x ^ { 2 } - 3 x + 2 }} \\ {\frac{4 x^{2}-4 x+3}{0+x-1}}\end{array}$

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

$4 x^{2}-4 x+3=0 \Rightarrow$

$\begin{array}{l}{a=4} \\ {b=-4} \\ {c=3}\end{array}$

$b^{2}-4 a c=(4)^{2}-4(4)(3)$

$=16-48=-32<0$

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

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

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

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

let $u=x-\frac{1}{2} \rightarrow d u=1 d x$

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

$=\frac{1}{4}\left[\int \frac{u}{u^{2}+\frac{1}{2}}dx-\int \frac{\frac{1}{2}}{u^{2}+\frac{1}{2}} dx\right]$

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

$\int {1} d x = {x+c_{1}}$

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

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

$u=x-\frac{1}{2}$

$d u=d x$

$=\frac{1}{4}\left[\int \frac{u}{u^{2}+\frac{1}{2}} d x-\frac{1}{2} \int \frac{1}{u^{2}+\frac{1}{2}} d x\right]$

$=\frac{1}{4}\left[\frac{1}{2} \int \frac{2 u}{u^{2}+\frac{1}{2}} d x-\frac{1}{2} \int \frac{1}{u^{2}+\frac{1}{2}} d x\right]$

$=\frac{1}{4}\left[\frac{1}{2} \int \frac{2 u}{u^{2}+\frac{1}{2}} d u-\frac{1}{2} \int \frac{1}{u^{2}+\frac{1}{2}} d u\right]$

$=\frac{1}{4}\left[\frac{1}{2} \cdot \ln \left|u^{2}+\frac{1}{2}\right|+c-\frac{1}{2} \cdot \sqrt{2} \tan ^{-1}(\sqrt{2} \cdot u)+c\right]$

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

$=x+\frac{1}{4}\left[\frac{1}{2} \cdot \ln \left|u^{2}+\frac{1}{2}\right|-\frac{1}{2}\cdot \sqrt{2} \tan ^{-1}(\sqrt{2} \cdot u)\right]+c$

$\int f(x)=x+\frac{1}{8} \ln \left|\left(x-\frac{1}{2}\right)^{2}+\frac{1}{2}\right|-\frac{1}{4 \sqrt{2}} \tan ^{-1}\left(\sqrt{2} \cdot\left(x-\frac{1}{2}\right)\right]+c$