Ta có :

\(\frac{3x+2}{x^2+2x-3}=\frac{E\left(2x+2\right)+D}{x^2+2x-3}=\frac{2E+D+2E}{x^2+2x-3}\)

Đồng nhất hệ số hai tử sốta có hệ phương trình 

\(\begin{cases}2E=3\\D+2E=2\end{cases}\) \(\Rightarrow\begin{cases}E=\frac{3}{2}\\D=-1\end{cases}\)

\(\Rightarrow\) \(\frac{3x+2}{x^2+2x-3}=\frac{\frac{3}{2}\left(2x+2\right)}{x^2+2x-3}-\frac{1}{x^2+2x-3}\)

Vậy :

\(\int\frac{3x+2}{x^2+2x-3}dx=\frac{3}{2}\int\frac{d\left(x^2+2x-3\right)}{x^2+2x-3}+\int\frac{1}{x^2+2x-3}dx\)\(=\frac{3}{2}\ln\left|x^2+2x-3\right|+J\left(1\right)\)

Tính :

\(J=\int\frac{1}{x^2+2x-3}dx=\frac{1}{4}\left(\int\frac{1}{x-1}dx-\int\frac{1}{x+3}dx\right)=\frac{1}{4}\ln\left|x-1\right|-\ln\left|x+3\right|=\frac{1}{4}\ln\left|\frac{x-1}{x+3}+C\right|\)

Do đó :  \(\int\frac{3x+2}{x^2+2x-3}dx=\frac{3}{2}\ln\left|x^2+2x-3\right|+\frac{1}{4}\ln\left|\frac{x-1}{x+3}\right|+C\)

b) Ta có :

\(\frac{2x-3}{x^2+4x+4}=\frac{E\left(2x+4\right)+D}{x^2+4x+4}=\frac{2Ex+D+4E}{x^2+4x+4}\)

Đồng nhất hệ số  hai tử số :

Ta có hệ : \(\Leftrightarrow\)\(\begin{cases}2E=2\\D+4E=-3\end{cases}\)\(\Leftrightarrow\)\(\begin{cases}E=1\\D=-7\end{cases}\)

Suy ra :

\(\frac{2x-3}{x^2+4x+4}=\frac{2x+4}{x^2+4x+4}-\frac{7}{x^2+4x+4}\)

Vậy : \(\int\frac{2x-3}{x^2+4x+4}dx=\int\frac{2x+4}{x^2+4x+4}dx-7\int\frac{1}{\left(x+2\right)^2}dx=\ln\left|x^2+4x+4\right|+\frac{7}{x+2}+C\)

a) \(\int\frac{1}{x^2-3x+2}dx=\frac{1}{2-1}\int\frac{1}{\left(x-1\right)\left(x-2\right)}dx\)

=\(\int\frac{1}{x-2}dx-\int\frac{1}{x-1}dx=ln\left|x-2\right|-ln\left|x-1\right|=ln\left|\frac{x-2}{x-1}+C\right|\)

b) \(\int\frac{1}{4x^2-3x-1}dx=\frac{1}{4}.\frac{1}{\left(1-\frac{1}{4}\right)}\int\frac{1}{\left(x+\frac{1}{4}\right)\left(x-1\right)}dx\)

=\(\frac{1}{3}.\left[\int\frac{1}{x-1}dx-\int\frac{1}{x+\frac{1}{4}}dx\right]\)

=\(\frac{1}{3}\left[ln\left|x-1\right|-ln\left|x+\frac{1}{4}\right|\right]=\frac{1}{3}ln\left|\frac{x-1}{x+\frac{1}{4}}\right|+C\)

=\(\frac{1}{3}ln\left|\frac{4\left(x-1\right)}{4x+1}+C\right|\)

\(\dfrac{x^2-4x+2}{x^2+2x-3}\)

\(=\dfrac{x^2+2x-3-6x-5}{x^2+2x-3}\)

\(=1-\dfrac{6x+5}{\left(x+3\right)\left(x-1\right)}\)

Đặt \(\dfrac{6x+5}{\left(x+3\right)\left(x-1\right)}=\dfrac{A}{x+3}+\dfrac{B}{x-1}\)

=>\(6x+5=A\left(x-1\right)+B\left(x+3\right)\)

=>\(6x+5=x\left(A+B\right)-A+3B\)

=>\(\left\{{}\begin{matrix}A+B=6\\-A+3B=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}B=\dfrac{11}{4}\\A=6-\dfrac{11}{4}=\dfrac{13}{4}\end{matrix}\right.\)

vậy: \(\dfrac{x^2-4x+2}{x^2+2x-3}=1-\dfrac{13}{4x+12}-\dfrac{11}{4x-4}\)

\(\int\dfrac{x^2-4x+2}{x^2+2x-3}dx=\int1-\dfrac{13}{4x+12}-\dfrac{11}{4x-4}dx\)

\(=x-\dfrac{13}{4}\cdot ln\left|x+3\right|-\dfrac{11}{4}\cdot ln\left|x-1\right|\)

(\(=\left(\frac{1}{20}x^5+\frac{4}{3}x^4+\frac{3}{5}x^{\frac{3}{5}}+\frac{1}{2}x^2-2x+C\right)\)

a)

\(\int\frac{2\left(x_{ }+1\right)}{x^2+2x_{ }-3}dx=\int\frac{2x+2}{x^2+2x-3}dx\)

\(=\int\frac{d\left(x^2+2x-3\right)}{x^2+2x-3}=ln\left|x^2+2x-3\right|+C\)

b)\(\int\frac{2\left(x-2\right)dx}{x^2-4x+3}=\int\frac{2x-4dx}{x^2-4x+3}=\int\frac{d\left(x^2-4x+3\right)}{x^2-4x+3}=ln\left|x^2-4x+3\right|+C\)

a)

\(\frac{1}{x^2-4x+4}dx=\frac{1}{\left(x-2\right)^2}dx=-\frac{1}{x-2}+C\)

b) \(\frac{1}{9x^2-12x+4}dx=\frac{1}{9\left(x-\frac{2}{3}\right)^2}dx=\frac{1}{9}.\frac{1}{\left(x-\frac{2}{3}\right)^2}dx=\frac{1}{9}.\frac{1}{x-\frac{2}{3}}=\frac{1}{9x-6}+C\)

a) Theo công thức 3) trong bảng nguyên hàm ta có :

\(\int3^x5^{2x}dx=\int3^x\left(25\right)^xdx=\int\left(75\right)^xdx=\frac{75^x}{\ln75}+C\)

b) Áp dụng các công thức I, II ( định lí 4.2) và 2), 3) trong bảng nguyên hàm ta có 

\(\int\left(x^2+2e^x\right)dx=\int x^{2^{ }}dx+2\int e^xdx=\frac{1}{3}x^3+2e^x+C\)

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

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

                     \(=\frac{x^3}{3}+x+\frac{1}{2}\ln\left|\frac{x-1}{x+1}\right|+C\)

d) Nhân tử số và mẫu số của biểu thức dưới dấu nguyên hàm với biểu thức liên hợp với mẫu số ta thu được.

\(\int\frac{dx}{\sqrt{4x+1}+\sqrt{4x-2}}=\int\frac{\sqrt{4x+1}-\sqrt{4x-2}}{3}dx\)

                         \(=\frac{1}{3.4}\int\left(4x+1\right)^{\frac{1}{2}}d\left(4x+1\right)-\frac{1}{3.4}\int\left(4x-2\right)^{\frac{1}{2}}d\left(4x-2\right)\)

                         \(=\frac{1}{12}\left[\sqrt{\left(4x+1\right)^3}-\sqrt{\left(4x-2\right)^3}\right]+C\)

\(=\int\left(6x^2-\dfrac{4}{x}+sin3x-cos4x+e^{2x+1}+9^{x-1}+\dfrac{1}{cos^2x}-\dfrac{1}{sin^2x}\right)dx\)

\(=2x^3-4ln\left|x\right|-\dfrac{1}{3}cos3x-\dfrac{1}{4}sin4x+\dfrac{1}{2}e^{2x+1}+\dfrac{9^{x-1}}{ln9}+tanx+cotx+C\)

\(=\left(\frac{m}{4}x^4-x^3+\frac{2}{3}\left(x-1\right)^{\frac{3}{2}}-\frac{4m}{2.x^2}-\frac{5}{2.x^2}-7mx+C\right)\)