4 câu 1,3,4,5 giống nhau, mình làm 1 câu và bạn dựa vào đó tự xử lý mấy câu còn lại nhé

1/ \(I=\int sin2x.e^{3x}dx\) \(\Rightarrow\left\{{}\begin{matrix}u=sin2x\\dv=e^{3x}dx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=2cos2x.dx\\v=\dfrac{1}{3}e^{3x}\end{matrix}\right.\)

\(\Rightarrow I=\dfrac{1}{3}sin2x.e^{3x}-\dfrac{2}{3}\int cos2x.e^{3x}dx=\dfrac{1}{3}sin2x.e^{3x}-\dfrac{2}{3}I_1\)

Xét \(I_1=\int cos2x.e^{3x}dx\) \(\Rightarrow\left\{{}\begin{matrix}u=cos2x\\dv=e^{3x}dx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=-2sin2xdx\\v=\dfrac{1}{3}e^{3x}\end{matrix}\right.\)

\(\Rightarrow I_1=\dfrac{1}{3}cos2x.e^{3x}+\dfrac{2}{3}\int sin2x.e^{3x}dx=\dfrac{1}{3}cos2x.e^{3x}+\dfrac{2}{3}I\)

\(\Rightarrow I=\dfrac{1}{3}sin2x.e^{3x}-\dfrac{2}{3}\left(\dfrac{1}{3}cos2x.e^{3x}+\dfrac{2}{3}I\right)\)

\(\Rightarrow\dfrac{13}{9}I=\dfrac{1}{9}e^{3x}\left(3sin2x-2cos2x\right)\)

\(\Rightarrow I=\dfrac{1}{13}e^{3x}\left(3sin2x-2cos2x\right)+C\)

3/ \(\int e^x\left(\dfrac{1+cos2x}{2}\right)dx=\dfrac{1}{2}\int e^xdx+\dfrac{1}{2}\int cos2x.e^xdx=\dfrac{e^x}{2}+\dfrac{1}{2}I_1\)

\(I_1\) có cách tính y hệt như bài 1, bạn nguyên hàm từng phần 2 lần là xong

4/ Cũng hạ bậc tương tự câu trên và xử lý

5/ \(I=\int e^{-x}\left(\dfrac{cos3x+3cosx}{4}\right)dx=\dfrac{1}{4}\int e^{-x}\left(cos3x+3cosx\right)dx\)

\(\Rightarrow I=\dfrac{1}{4}\int e^{-x}cos3x.dx+\dfrac{3}{4}\int e^{-x}cosx.dx=I_1+I_2\)

Dùng phương pháp tương tự bài 1, lần lượt tính \(I_1\) và \(I_2\) rồi cộng vào

2/\(I=\int\dfrac{x^4}{\left(x^2-1\right)^2}dx=\int\left(1+\dfrac{2x^2-1}{\left(x^2-1\right)^2}\right)dx=\int\left(1+\dfrac{2}{x^2-1}+\dfrac{1}{\left(x^2-1\right)^2}\right)dx\)

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

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

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

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

\(=x+\dfrac{3}{4}ln\left|\dfrac{x-1}{x+1}\right|-\dfrac{x}{2\left(x^2-1\right)}+C\)

\(I=\int\dfrac{x}{1-cos2x}dx=\int\dfrac{x}{2sin^2x}dx\)

Đặt \(\left\{{}\begin{matrix}u=\dfrac{x}{2}\\dv=\dfrac{1}{sin^2x}dx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=\dfrac{dx}{2}\\v=-cotx\end{matrix}\right.\)

\(\Rightarrow I=\dfrac{-x.cotx}{2}+\dfrac{1}{2}\int cotxdx=\dfrac{-x.cotx}{2}+\dfrac{1}{2}\int\dfrac{cosx.dx}{sinx}\)

\(=\dfrac{-x.cotx}{2}+\dfrac{1}{2}\int\dfrac{d\left(sinx\right)}{sinx}=\dfrac{-x.cotx}{2}+\dfrac{1}{2}ln\left|sinx\right|+C\)

2/ Câu 2 bữa trước làm rồi, bạn coi lại nhé

3/ \(I=\int\left(2x+1\right)ln^2xdx\)

Đặt \(\left\{{}\begin{matrix}u=ln^2x\\dv=\left(2x+1\right)dx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=\dfrac{2lnx}{x}dx\\v=x^2+x\end{matrix}\right.\)

\(\Rightarrow I=\left(x^2+x\right)ln^2x-\int\left(2x+2\right)lnxdx=\left(x^2+x\right)ln^2x-I_1\)

\(I_1=\int\left(2x+2\right)lnx.dx\) \(\Rightarrow\left\{{}\begin{matrix}u=lnx\\dv=\left(2x+2\right)dx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=\dfrac{dx}{x}\\v=x^2+2x\end{matrix}\right.\)

\(\Rightarrow I_1=\left(x^2+2x\right)lnx-\int\left(x+2\right)dx=\left(x^2+2x\right)ln-\dfrac{x^2}{2}+2x+C\)

\(\Rightarrow I=\left(x^2+x\right)ln^2x-\left(x^2+2x\right)lnx+\dfrac{x^2}{2}-2x+C\)

4/ \(I=\int\left(2x-1\right)cosx.dx\) \(\Rightarrow\left\{{}\begin{matrix}u=2x-1\\dv=cosx.dx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=2dx\\v=sinx\end{matrix}\right.\)

\(\Rightarrow I=\left(2x-1\right)sinx-2\int sinx.dx=\left(2x-1\right)sinx+2cosx+C\)

5/ \(I=\int\left(x^2+x+1\right)e^xdx\) \(\Rightarrow\left\{{}\begin{matrix}u=x^2+x+1\\dv=e^xdx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=\left(2x+1\right)dx\\v=e^x\end{matrix}\right.\)

\(\Rightarrow I=\left(x^2+x+1\right)e^x-\int\left(2x+1\right)e^xdx\)

\(I_1=\int\left(2x+1\right)e^xdx\) \(\Rightarrow\left\{{}\begin{matrix}u=2x+1\\dv=e^xdx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=2dx\\v=e^x\end{matrix}\right.\)

\(\Rightarrow I_1=\left(2x+1\right)e^x-2\int e^xdx=\left(2x+1\right)e^x-2e^x+C=\left(2x-1\right)e^x+C\)

\(\Rightarrow I=\left(x^2+x+1\right)e^x-\left(2x-1\right)e^x+C=\left(x^2-x+2\right)e^x+C\)

6/ \(I=\int\left(2x+1\right).ln\left(x+2\right)dx\)

\(\Rightarrow\left\{{}\begin{matrix}u=ln\left(x+2\right)\\dv=\left(2x+1\right)dx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=\dfrac{dx}{x+2}\\v=x^2+x\end{matrix}\right.\)

\(\Rightarrow I=\left(x^2+x\right)ln\left(x+2\right)-\int\dfrac{x^2+x}{x+2}dx\)

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

\(I=\left(x^2+x\right)ln\left(x+2\right)-\dfrac{x^2}{2}+x-2ln\left|x+2\right|+C\)

Khi \(x=4>3\Rightarrow f\left(x\right)=2x-3\)

\(\Rightarrow f\left(4\right)=2.4-3=5\)

Câu 2)

Đặt \(\left\{\begin{matrix} u=\ln ^2x\\ dv=x^2dx\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=2\frac{\ln x}{x}dx\\ v=\frac{x^3}{3}\end{matrix}\right.\Rightarrow I=\frac{x^3}{3}\ln ^2x-\frac{2}{3}\int x^2\ln xdx\)

Đặt \(\left\{\begin{matrix} k=\ln x\\ dt=x^2dx\end{matrix}\right.\Rightarrow \left\{\begin{matrix} dk=\frac{dx}{x}\\ t=\frac{x^3}{3}\end{matrix}\right.\Rightarrow \int x^2\ln xdx=\frac{x^3\ln x}{3}-\int \frac{x^2}{3}dx=\frac{x^3\ln x}{3}-\frac{x^3}{9}+c\)

Do đó \(I=\frac{x^3\ln^2x}{3}-\frac{2}{9}x^3\ln x+\frac{2}{27}x^3+c\)

Câu 3:

\(I=\int\frac{2}{\cos 2x-7}dx=-\int\frac{2}{2\sin^2x+6}dx=-\int\frac{dx}{\sin^2x+3}\)

Đặt \(t=\tan\frac{x}{2}\Rightarrow \left\{\begin{matrix} \sin x=\frac{2t}{t^2+1}\\ dx=\frac{2dt}{t^2+1}\end{matrix}\right.\)

\(\Rightarrow I=-\int \frac{2dt}{(t^2+1)\left ( \frac{4t^2}{(t^2+1)^2}+3 \right )}=-\int\frac{2(t^2+1)dt}{3t^4+10t^2+3}=-\int \frac{2d\left ( t-\frac{1}{t} \right )}{3\left ( t-\frac{1}{t} \right )^2+16}=\int\frac{2dk}{3k^2+16}\)

Đặt \(k=\frac{4}{\sqrt{3}}\tan v\). Đến đây dễ dàng suy ra \(I=\frac{-1}{2\sqrt{3}}v+c\)

Câu 6)

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

Câu 8)

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

\(\Leftrightarrow I=\int \ln (x+1)d(x+1)-\int \ln (x-1)d(x-1)\)

Xét \(\int \ln tdt\) ta có:

Đặt \(\left\{\begin{matrix} u=\ln t\\ dv=dt\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=\frac{dt}{t}\\ v=t\end{matrix}\right.\Rightarrow \int \ln tdt=t\ln t-\int dt=t\ln t-t+c\)

\(\Rightarrow I=(x+1)\ln (x+1)-(x+1)-(x-1)\ln (x-1)+x-1+c\)

\(\Leftrightarrow I=(x+1)\ln(x+1)-(x-1)\ln(x-1)+c\)

1/ \(I=\int\limits^1_0\dfrac{2x+1}{x^2+x+1}dx=\int\limits^1_0\dfrac{d\left(x^2+x+1\right)}{x^2+x+1}=ln\left|x^2+x+1\right||^1_0=ln3\)

2/ \(\int\limits^{\dfrac{1}{2}}_0\dfrac{5x}{\left(1-x^2\right)^3}dx=-\dfrac{5}{2}\int\limits^{\dfrac{1}{2}}_0\dfrac{d\left(1-x^2\right)}{\left(1-x^2\right)^3}=\dfrac{5}{4}\dfrac{1}{\left(1-x^2\right)^2}|^{\dfrac{1}{2}}_0=\dfrac{35}{36}\)

3/ \(\int\limits^1_0\dfrac{2x}{\left(x+1\right)^3}dx\Rightarrow\) đặt \(x+1=t\Rightarrow x=t-1\Rightarrow dx=dt;\left\{{}\begin{matrix}x=0\Rightarrow t=1\\x=1\Rightarrow t=2\end{matrix}\right.\)

\(I=\int\limits^2_1\dfrac{2\left(t-1\right)dt}{t^3}=\int\limits^2_1\left(\dfrac{2}{t^2}-\dfrac{2}{t^3}\right)dt=\left(\dfrac{-2}{t}+\dfrac{1}{t^2}\right)|^2_1=\dfrac{1}{4}\)

4/ \(\int\limits^1_0\dfrac{4x-2}{\left(x^2+1\right)\left(x+2\right)}dx\)

Kĩ thuật chung là tách và sử dụng hệ số bất định như sau:

\(\dfrac{4x-2}{\left(x^2+1\right)\left(x+2\right)}=\dfrac{ax+b}{x^2+1}+\dfrac{c}{x+2}=\dfrac{\left(a+c\right)x^2+\left(2a+b\right)x+2b+c}{\left(x^2+1\right)\left(x+2\right)}\)

\(\Rightarrow\left\{{}\begin{matrix}a+c=0\\2a+b=4\\2b+c=-2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}b=0\\a=-c=2\end{matrix}\right.\)

\(\Rightarrow I=\int\limits^1_0\left(\dfrac{2x}{x^2+1}-\dfrac{2}{x+2}\right)dx=\int\limits^1_0\dfrac{d\left(x^2+1\right)}{x^2+1}-2\int\limits^1_0\dfrac{d\left(x+2\right)}{x+2}=ln\dfrac{8}{9}\)

5/ \(\int\limits^1_0\dfrac{x^2dx}{x^6-9}\Rightarrow\) đặt \(x^3=t\Rightarrow3x^2dx=dt\Rightarrow x^2dx=\dfrac{1}{3}dt;\left\{{}\begin{matrix}x=0\Rightarrow t=0\\x=1\Rightarrow t=1\end{matrix}\right.\)

\(I=\dfrac{1}{3}\int\limits^1_0\dfrac{dt}{t^2-9}=\dfrac{1}{18}\int\limits^1_0\left(\dfrac{1}{t-3}-\dfrac{1}{t+3}\right)dt=\dfrac{1}{18}ln\left|\dfrac{t-3}{t+3}\right||^1_0=-\dfrac{1}{18}ln2\)

6/ Tương tự câu 4, sử dụng hệ số bất định ta tách được:

\(\int\limits^2_1\dfrac{2x-1}{x^2\left(x+1\right)}dx=\int\limits^2_1\left(\dfrac{3x-1}{x^2}-\dfrac{3}{x+1}\right)dx=\int\limits^2_1\left(\dfrac{3}{x}-\dfrac{1}{x^2}-\dfrac{3}{x+1}\right)dx\)

\(=\left(3ln\left|\dfrac{x}{x+1}\right|+\dfrac{1}{x}\right)|^2_1=3ln\dfrac{4}{3}-\dfrac{1}{2}\)

Câu 1)

\(I=\int \ln ^3 xdx\). Đặt \(\left\{\begin{matrix} u=\ln ^3x\\ dv=dx\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=\frac{3\ln ^2x}{x}dx\\ v=x\end{matrix}\right.\)

\(\Rightarrow I=x\ln ^3x-3\int \ln^2xdx\)

Tiếp tục nguyên hàm từng phần cho \(\int \ln ^2xdx\) như trên, ta suy ra:

\(\int\ln ^2xdx=x\ln^2x-2\int \ln x dx\).

Tiếp tục nguyên hàm từng phần cho \(\int \ln xdx\Rightarrow \int \ln xdx=x\ln x-x+c\)

Do đó mà \(I=x\ln ^3x-3(x\ln^2x-2x\ln x+2x)+c\)

\(\Leftrightarrow I=x\ln^3x-3x\ln^2x+6x\ln x-6x+c\)

Câu 2)

\(I=\int ^{1}_{0}(x+\sin ^2x)\cos x dx=\int ^{1}_{0}x\cos xdx+\int ^{1}_{0}\sin^2x\cos xdx\)

Đặt \(\left\{\begin{matrix} u=x\\ dv=\cos xdx\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=dx\\ v=\sin x\end{matrix}\right.\Rightarrow \int x\cos xdx=x\sin x-\int \sin xdx=x\sin x+\cos x+c\)

\(\Rightarrow \int ^{1}_{0} x\cos xdx=\sin 1+\cos 1-1\)

Còn \(\int ^{1}_{0}\sin^2x\cos xdx=\int ^{1}_{0}\sin ^2xd(\sin x)=\left.\begin{matrix} 1\\ 0\end{matrix}\right|\frac{\sin ^3x}{3}=\frac{\sin^31}{3}\)

\(\Rightarrow I=-1+\sin 1+\cos 1+\frac{\sin ^3 1}{3}\approx 0,0173\)

Câu 3:

Đối với \(\int xe^{2x}dx\)

\(\left\{\begin{matrix} u=x\\ dv=e^{2x}dx\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=dx\\ v=\int e^{2x}dx=\frac{e^{2x}}{2}\end{matrix}\right.\)

\(\Rightarrow \int xe^{2x}=\frac{1}{2}xe^{2x}-\frac{1}{2}\int e^{2x}dx=\frac{1}{2}xe^{2x}-\frac{1}{4}e^{2x}+c\)

Đối với \(\int x\sqrt[3]{x+1}dx=\int \sqrt[3]{(x+1)^4}dx-\int \sqrt{x+1}dx=\frac{3(x+1)^\frac{7}{3}}{7}-\frac{3}{4}(x+1)^{\frac{4}{3}}+c\)

\(\Rightarrow \int x\sqrt[3]{x+1}dx=\frac{3(x+1)^{\frac{4}{3}}(4x-3)}{28}\)

Do đó mà \(\int x(e^{2x}-\sqrt[3]{x+1})dx=\frac{1}{2}xe^{2x}-\frac{1}{4}e^{2x}+\frac{3(x+1)^{\frac{4}{3}}(4x-3)}{28}+c\)

a)

\(f\left(-2\right)=\left(-2\right).\left(-2\right)+3=4+3=7\)

\(f\left(-1\right)=\left(-2\right).\left(-1\right)+3=2+3=5\)

\(f\left(0\right)=\left(-2\right).0+3=0+3=3\)

\(f\left(-\frac{1}{2}\right)=\left(-2\right).\left(-\frac{1}{2}\right)+3=1+3=4\)

\(f\left(\frac{1}{2}\right)=\left(-2\right).\frac{1}{2}+3=\left(-1\right)+3=2\)

Câu b thì bạn cứ thế số vào và làm tương tự vậy.

chúc bạn học tốt

b)g(-1)=(-1)²-1=1-1=0

g(0)=0²-1=0-1=-1

g(1)=1²-1=1-1=0

g(2)=2²-1=4-1=3

a) \(\int\left(x+\ln x\right)x^2\text{d}x=\int x^3\text{d}x+\int x^2\ln x\text{dx}\)

\(=\dfrac{x^4}{4}+\int x^2\ln x\text{dx}+C\) (*)

Để tính: \(\int x^2\ln x\text{dx}\) ta sử dụng công thức tính tích phân từng phần như sau:

Đặt \(\left\{{}\begin{matrix}u=\ln x\\v'=x^2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}u'=\dfrac{1}{x}\\v=\dfrac{1}{3}x^3\end{matrix}\right.\)

Suy ra:

\(\int x^2\ln x\text{dx}=\dfrac{1}{3}x^3\ln x-\dfrac{1}{3}\int x^2\text{dx}\)

\(=\dfrac{1}{3}x^3\ln x-\dfrac{1}{3}.\dfrac{1}{3}x^3\)

Thay vào (*) ta tính được nguyên hàm của hàm số đã cho bằng:

(*) \(=\dfrac{1}{3}x^3-\dfrac{1}{3}x^3\ln x+\dfrac{1}{9}x^3+C\)

\(=\dfrac{4}{9}x^3-\dfrac{1}{3}x^3\ln x+C\)

b) Đặt \(\left\{{}\begin{matrix}u=x+\sin^2x\\v'=\sin x\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}u'=1+2\sin x.\cos x\\v=-\cos x\end{matrix}\right.\)

Ta có:

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

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

\(=-\left(x+\sin^2x\right)\cos x+\sin x-2\int\cos^2x.d\left(\cos x\right)\)

\(=-\left(x+\sin^2x\right)\cos x+\sin x-2\dfrac{\cos^3x}{3}+C\)

c) Đặt \(\left\{{}\begin{matrix}u=x+e^x\\v'=e^{2x}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}u'=1+e^x\\v=\dfrac{1}{2}e^{2x}\end{matrix}\right.\)

Ta có:

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

\(=\dfrac{1}{2}\left(x+e^x\right)e^{2x}-\dfrac{1}{2}\int e^{2x}\text{dx}-\dfrac{1}{2}\int e^{3x}\text{dx}\)

\(=\dfrac{1}{2}\left(x+e^x\right)e^{2x}-\dfrac{1}{2}.\dfrac{1}{2}e^{2x}-\dfrac{1}{2}.\dfrac{1}{3}e^{3x}\)

\(=\dfrac{1}{2}xe^{2x}-\dfrac{1}{4}e^{2x}+\dfrac{1}{3}e^{3x}\)

a) Mẫu số chứa các biểu thức có nghiệm  thực và không có nghiệm thực.

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

Tay x=1 vào 2 tử, ta có : 2=2A, vậy A=1

Do đó (1) trở thành : 

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

Đồng nhất hệ số hai tử số, ta có hệ :

\(\begin{cases}B+1=1\\C-B=2\\1-C=-1\end{cases}\)\(\Leftrightarrow\)\(\begin{cases}B=0\\C=2\\A=1\end{cases}\)\(\Rightarrow\)

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

Vậy :

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

* Tính \(J=\int\frac{1}{x^2+1}dx.\)

Đặt \(\begin{cases}x=\tan t\rightarrow dx=\left(1+\tan^2t\right)dt\\1+x^2=1+\tan^2t\end{cases}\)

Cho nên :

\(\int\frac{1}{x^2+1}dx=\int\frac{1}{1+\tan^2t}\left(1+\tan^2t\right)dt=\int dt=t;do:x=\tan t\Rightarrow t=arc\tan x\)

Do đó, thay tích phân J vào (2), ta có : 

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

b) Ta phân tích 

\(f\left(x\right)=\frac{x^2+1}{\left(x-1\right)^3\left(x+3\right)}=\frac{A}{\left(x-1\right)^3}+\frac{B}{\left(x-1\right)^2}+\frac{C}{x-1}+\frac{D}{x+3}\)\(=\frac{A\left(x+3\right)+B\left(x-1\right)\left(x+3\right)+C\left(x-1\right)^2\left(x+3\right)+D\left(x-1\right)^3}{\left(x-1\right)^3\left(x+3\right)}\)

Thay x=1 và x=-3 vào hai tử số, ta được :

\(\begin{cases}x=1\rightarrow2=4A\rightarrow A=\frac{1}{2}\\x=-3\rightarrow10=-64D\rightarrow D=-\frac{5}{32}\end{cases}\)

Thay hai giá trị của A và D vào (*) và đồng nhất hệ số hai tử số, ta cso hệ hai phương trình :

\(\begin{cases}0=C+D\Rightarrow C=-D=\frac{5}{32}\\1=3A-3B+3C-D\Rightarrow B=\frac{3}{8}\end{cases}\)

\(\Rightarrow f\left(x\right)=\frac{1}{2\left(x-1\right)^3}+\frac{3}{8\left(x-1\right)^2}+\frac{5}{32\left(x-1\right)}-+\frac{5}{32\left(x+3\right)}\)

Vậy : 

\(\int\frac{x^2+1}{\left(x-1\right)^3\left(x+3\right)}dx=\)\(\left(\frac{1}{2\left(x-1\right)^3}+\frac{3}{8\left(x-1\right)^2}+\frac{5}{32\left(x-1\right)}-+\frac{5}{32\left(x+3\right)}\right)dx\)

\(=-\frac{1}{a\left(x-1\right)^2}-\frac{3}{8\left(x-1\right)}+\frac{5}{32}\ln\left|x-1\right|-\frac{5}{32}\ln\left|x+3\right|+C\)

\(=-\frac{1}{a\left(x-1\right)^2}-\frac{3}{8\left(x-1\right)}+\frac{5}{32}\ln\left|\frac{x-1}{x+3}\right|+C\)

cac cau nghi gi ve cau hoi nay