\displaystyle \textbf{Question 1: }~\int \frac{x^2+5x+2}{x+2}\,dx.
\displaystyle \text{Answer:}
\displaystyle \frac{x^2+5x+2}{x+2}=x+3-\frac{4}{x+2},\ \text{since }x^2+5x+2=(x+2)(x+3)-4.
\displaystyle \int \frac{x^2+5x+2}{x+2}\,dx=\int\left(x+3-\frac{4}{x+2}\right)dx.
\displaystyle =\int x\,dx+\int 3\,dx-4\int \frac{1}{x+2}\,dx.
\displaystyle =\frac{x^2}{2}+3x-4\log|x+2|+C.
\displaystyle \\

\displaystyle \textbf{Question 2: }~\int \frac{x^3}{x-2}\,dx.
\displaystyle \text{Answer:}
\displaystyle \frac{x^3}{x-2}=x^2+2x+4+\frac{8}{x-2},\ \text{since }x^3=(x-2)(x^2+2x+4)+8.
\displaystyle \int \frac{x^3}{x-2}\,dx=\int\left(x^2+2x+4+\frac{8}{x-2}\right)dx.
\displaystyle =\int x^2dx+2\int xdx+4\int dx+8\int \frac{1}{x-2}dx.
\displaystyle =\frac{x^3}{3}+x^2+4x+8\log|x-2|+C.
\displaystyle \\

\displaystyle \textbf{Question 3: }~\int \frac{x^2+x+5}{3x+2}\,dx.
\displaystyle \text{Answer:}
\displaystyle \frac{x^2+x+5}{3x+2}=Ax+B+\frac{C}{3x+2}.
\displaystyle x^2+x+5=(Ax+B)(3x+2)+C.
\displaystyle (Ax+B)(3x+2)=3Ax^2+(2A+3B)x+2B.
\displaystyle 3A=1,\ 2A+3B=1,\ 2B+C=5.
\displaystyle A=\frac{1}{3},\ B=\frac{1}{9},\ C=\frac{43}{9}.
\displaystyle \frac{x^2+x+5}{3x+2}=\frac{1}{3}x+\frac{1}{9}+\frac{43/9}{3x+2}.
\displaystyle \int \frac{x^2+x+5}{3x+2}\,dx=\int\left(\frac{1}{3}x+\frac{1}{9}+\frac{43/9}{3x+2}\right)dx.
\displaystyle =\frac{x^2}{6}+\frac{x}{9}+\frac{43}{27}\log|3x+2|+C.
\displaystyle \\

\displaystyle \textbf{Question 4: }~\int \frac{2x+3}{(x-1)^2}\,dx.
\displaystyle \text{Answer:}
\displaystyle x-1=t,\ x=t+1,\ dx=dt.
\displaystyle \int \frac{2x+3}{(x-1)^2}\,dx=\int \frac{2(t+1)+3}{t^2}dt=\int\left(\frac{2}{t}+\frac{5}{t^2}\right)dt.
\displaystyle =2\log|t|-\frac{5}{t}+C.
\displaystyle =2\log|x-1|-\frac{5}{x-1}+C.
\displaystyle \\

\displaystyle \textbf{Question 5: }~\int \frac{x^2+3x-1}{(x+1)^2}\,dx.
\displaystyle \text{Answer:}
\displaystyle x+1=t,\ x=t-1,\ dx=dt.
\displaystyle \int \frac{x^2+3x-1}{(x+1)^2}\,dx=\int \frac{(t-1)^2+3(t-1)-1}{t^2}dt.
\displaystyle =\int \frac{t^2+t-3}{t^2}dt=\int\left(1+\frac{1}{t}-\frac{3}{t^2}\right)dt.
\displaystyle =t+\log|t|+\frac{3}{t}+C.
\displaystyle =x+\log|x+1|+\frac{3}{x+1}+C.
\displaystyle \\

\displaystyle \textbf{Question 6: }~\int \frac{2x-1}{(x-1)^2}\,dx.
\displaystyle \text{Answer:}
\displaystyle x-1=t,\ x=t+1,\ dx=dt.
\displaystyle \int \frac{2x-1}{(x-1)^2}\,dx=\int \frac{2(t+1)-1}{t^2}dt=\int\left(\frac{2}{t}+\frac{1}{t^2}\right)dt.
\displaystyle =2\log|t|-\frac{1}{t}+C.
\displaystyle =2\log|x-1|-\frac{1}{x-1}+C.
\displaystyle \\

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