Đặt \(u=\left(x^3-2x^x+3x+1\right)\Rightarrow du=\left(3x^2-4x+3\right)dx;dv=\frac{dx}{e^{2x}}\Rightarrow v=-\frac{2}{e^{2x}}\)

Ta được : \(-\frac{2}{e^{2x}}\left(x^3-2x^2+3x+1\right)|^1_0+2\int\limits^1_0\left(\frac{3x^2-4x+3}{e^{2x}}\right)dx=2-\frac{6}{e^2}+2J\)

Tương tự ta tính J

Đăth \(u_1=\left(3x^2-4x+3\right)\Rightarrow du_1=\left(6x-4\right)dx;dv_1=\frac{dx}{e^{2x}}\Rightarrow v_1=-\frac{2}{e^{2x}}\left(1\right)\)

Do đó :

\(J=-\frac{2}{e^{2x}}\left(3x^2-4x+3\right)|^1_0+2\int\limits^1_0\frac{6x-4}{e^{2x}}dx=6-\frac{4}{e^2}+2K\left(2\right)\)

Ta tính K :

\(K=\int\limits^1_0\frac{6x-4}{e^{2x}}dx\)

Đặt \(u_2=6x-4\Rightarrow du_2=6dx;dv_2=\frac{dx}{e^{2x}}\Rightarrow v_2=-\frac{2}{e^{2x}}\)

Do đó : \(K=-\frac{2}{e^{2x}}\left(x-4\right)|^1_0+2\int\limits^1_0\frac{6dx}{e^{2x}}=\frac{6}{e^x}-8-6\frac{1}{e^{2x}}|^1_0\left(\frac{1}{e^2}-1\right)=-2\left(3\right)\)

Thay (3) vào (2) 

\(J=6-\frac{4}{e^2}+2\left(-2\right)=2-\frac{4}{e^2}\)

Lại thay vào (1) ta có :

\(I=2-\frac{6}{e^2}+2\left(2-\frac{4}{e^2}\right)=6-\frac{14}{e^2}\)

a) Đặt \(x=\sin t;t\in\left(-\frac{\pi}{2};\frac{\pi}{2}\right)\) \(\Rightarrow dx=\cos tdt\)

Suy ra : \(\frac{dx}{\sqrt{\left(1-x^2\right)^3}}=\frac{\cos tdt}{\sqrt{\left(1-\sin^2t\right)^3}}=\frac{\cos tdt}{\cos^3t}=\frac{dt}{\cos^2t}=d\left(\tan t\right)\)

Khi đó \(\int\frac{dx}{\sqrt{\left(1-x^2\right)^3}}=\int d\left(\tan t\right)=\tan t+C=\frac{\sin t}{\sqrt{1-\sin^2t}}=\frac{x}{\sqrt{1-x^2}}+C\)

b) Vì \(x^2+2x+3=\left(x+1\right)^2+\left(\sqrt{2}\right)^2\)

nên ta đặt : \(x+1=\sqrt{2}\tan t;t\in\left(-\frac{\pi}{2};\frac{\pi}{2}\right)\Rightarrow dx=\sqrt{2}.\frac{dt}{\cos^2t};\tan t=\frac{x+1}{\sqrt{2}}\)

Suy ra \(\frac{dx}{\sqrt{x^2+2x+3}}=\frac{dx}{\sqrt{\left(x^2+1\right)^2+\left(\sqrt{2}\right)^2}}=\frac{dx}{\sqrt{2\left(\tan^2t+1\right).\cos^2t}}\)

                        \(=\frac{dt}{\sqrt{2}\cos t}=\frac{1}{\sqrt{2}}.\frac{\cos tdt}{1-\sin^2t}=-\frac{1}{2\sqrt{2}}.\left(\frac{\cos tdt}{\sin t-1}-\frac{\cos tdt}{\sin t+1}\right)\)

Khi đó \(\int\frac{dx}{\sqrt{x^2+2x+3}}=-\frac{1}{2\sqrt{2}}\int\left(\frac{\cos tdt}{\sin t-1}-\frac{\cos tdt}{\sin t+1}\right)=-\frac{1}{2\sqrt{2}}\ln\left|\frac{\sin t-1}{\sin t+1}\right|+C\left(1\right)\)

Từ \(\tan t=\frac{x+1}{\sqrt{2}}\Leftrightarrow\tan^2t=\frac{\sin^2t}{1-\sin^2t}=\frac{\left(x+1\right)^2}{2}\Rightarrow\sin^2t=1-\frac{2}{x^2+2x+3}\)

Ta tìm được \(\sin t\) thay vào (1), ta tính được I

TA có

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

=\(\int\frac{1}{x}dx+\frac{5}{3}\int\left(\frac{1}{x-3}-\frac{1}{x}\right)dx=-\frac{2}{3}\int\frac{1}{x}dx+\frac{5}{3}\int\frac{1}{x-3}dx=\frac{-2}{3}ln\left|x\right|+\frac{5}{3}ln\left|x-3\right|+C\)

a)

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

Đặt

\(\left(x-\frac{1}{4}\right)=\frac{\sqrt{3}}{2}tant\) => dx=\(\frac{\sqrt{3}}{2}\left(1+tan^2t\right)dt\) =>\(\frac{1}{x^2+x+1}dx=\frac{1}{\frac{3}{4}\left(1+tan^2t\right)+\frac{3}{4}}\left(1+tan^2t\right)dt=\frac{3}{4}dt=\frac{3}{4}t+C\) 

Với \(\left(x-\frac{1}{4}\right)=\frac{\sqrt{3}}{2}tant=>t=\left(\frac{2\sqrt{3}}{4x-1}\right)\)

Câu b nhá :

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

Đặt

 \(x+1=\sqrt{2}tant=>dx=\sqrt{2}\left(1+tan^2t\right)dt\)

=> \(\frac{1}{x^2+2x+3}dx=\frac{1}{2\left(tan^2t+1\right)}.\left(1+tan^2t\right)dt=\frac{1}{2}dt=\frac{1}{2}t+C\)

Với

\(x+1=\sqrt{2}tant=>tant=\frac{x+1}{\sqrt{2}}<=>t=arctan\left(\frac{x+1}{\sqrt{2}}\right)\)

\(I=\int\limits^{\frac{\pi}{2}}_0\frac{\sin x}{\cos2x+3\cos x+2}dx=\int\limits^{\frac{\pi}{2}}_0\frac{\sin x}{2\cos^2x+3\cos x+1}dx\)

Đặt \(\cos x=t\Rightarrow dt=-\sin dx\)

Với \(x=0\Rightarrow t=1\)

Với \(x=\frac{\pi}{2}\Rightarrow t=0\)

\(I=\int\limits^1_0\frac{dt}{2t^2+3t+1}=\int\limits^1_0\frac{dt}{\left(2t+1\right)\left(t+1\right)}=2\int\limits^1_0\left(\frac{1}{2t+1}+\frac{1}{2t+1}\right)dt\)

  \(=\left(\ln\frac{2t+1}{2t+1}\right)|^1_0=\ln\frac{3}{2}\)

Câu 1:

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

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

Đặt \(x-\frac{1}{x}=a,x+\frac{1}{x}=b\Rightarrow A=\frac{1}{2}\int \frac{da}{a^2+2}+\frac{1}{2}\int \frac{db}{b^2-2}\)

Bằng cách đặt \(a=\sqrt{2}\tan u (-\frac{\pi}{2}< u<\frac{\pi}{2})\)

\(\Rightarrow \frac{1}{2}\int \frac{da}{a^2+2}=\frac{\sqrt{2}}{4}\tan^{-1}\left (\frac{a}{\sqrt{2}} \right)+c\)

\(\frac{1}{2}\int \frac{db}{b^2-2}=\frac{1}{4\sqrt{2}}\int \left (\frac{1}{b-\sqrt{2}}-\frac{1}{b+\sqrt{2}} \right)db\)\(=\frac{1}{4\sqrt{2}}\ln|\frac{b-\sqrt{2}}{b+\sqrt{2}}|+c\)

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

Awn, chúc mừng năm mới!

Câu 2:

\(B=\int \frac{x^4+1}{x^6+1}=\int\frac{(x^2+1)^2-2x^2}{(x^2+1)(x^4-x^2+1)}dx=\int\frac{x^2+1}{x^4-x^2+1}dx-2\int \frac{x^2dx}{(x^3)^2+1}\)

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

Đặt \(x-\frac{1}{x}=a, x^3=b\). Cần tính \(B=\int\frac{da}{a^2+1}-\frac{2}{3}\int\frac{db}{b^2+1}\)

Đến đây bài toán trở về dạng quen thuộc . Đặt \(a=\tan u, b=\tan v\)

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

Câu 3:

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

Tính riêng lẻ từng phần :)

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

\(\int\frac{4dx}{x^3}=\frac{-2}{x^2}\)

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

Xong.

P/s: Đùa chứ bạn đào đâu ra toàn bài khoai @@

\(I=\int\limits^0_{\frac{-1}{2}}\frac{dx}{\left(x+1\right)\sqrt{3+2x-x^2}}=\int\limits^0_{\frac{-1}{2}}\frac{dx}{\left(x+1\right)\left(\sqrt{\left(x+1\right)\left(3-x\right)}\right)}\)

                                   \(=\int\limits^0_{\frac{-1}{2}}\frac{dx}{\left(x+1\right)^2\sqrt{\frac{3-x}{x+1}}}\)

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

Đổi cận : \(x=-\frac{1}{2}\Rightarrow t=\sqrt{7};x=0\Rightarrow t=\sqrt{3}\)

\(I=-\frac{1}{2}\int\limits^{\sqrt{3}}_{\sqrt{7}}dt=\frac{1}{2}\left(\sqrt{7}-\sqrt{3}\right)\)

1) Đặt \(t=1+\sqrt{x-1}\Leftrightarrow x=\left(t-1\right)^2+1\forall t\ge1\Rightarrow dx=d\left(t-1\right)^2=2dt\)

\(\Rightarrow I_1=\int\frac{\left(t-1\right)^2+1}{t}\cdot2dt=2\int\frac{t^2-2t+2}{t}dt=2\int\left(t-2+\frac{2}{t}\right)dt\\ =t^2-4t+4lnt+C\)

Thay x vào ta có...

2) \(I_2=\int\frac{2sinx\cdot cosx}{cos^3x-\left(1-cos^2x\right)-1}dx=\int\frac{-2cosx\cdot d\left(cosx\right)}{cos^3x+cos^2x-2}=\int\frac{-2t\cdot dt}{t^3+t-2}\)

\(I_2=\int\frac{-2t}{\left(t-1\right)\left(t^2+2t+2\right)}dt=-\frac{2}{5}\int\frac{dt}{t-1}+\frac{1}{5}\int\frac{2t+2}{t^2+2t+2}dt-\frac{6}{5}\int\frac{dt}{\left(t+1\right)^2+1}\)

Ta có:

\(\int\frac{2t+2}{t^2+2t+2}dt=\int\frac{d\left(t^2+2t+2\right)}{t^2+2t+2}=ln\left(t^2+2t+2\right)+C\)

\(\int\frac{dt}{\left(t+1\right)^2+1}=\int\frac{\frac{1}{cos^2m}}{tan^2m+1}dm=\int dm=m+C=arctan\left(t+1\right)+C\)

Thay x vào, ta có....

3)

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

\(I_3=\int\left(\frac{1}{2\sqrt{x}}+\frac{1}{2}+\frac{\sqrt{x+1}}{2\sqrt{x}}\right)dx=\sqrt{x}+\frac{x}{2}+\int\sqrt{\frac{x+1}{x}}\cdot\frac{dx}{2}\)

Xét \(\int\sqrt{\frac{x+1}{x}}\cdot\frac{dx}{2}\)

Đặt \(x=tan^2t\Leftrightarrow dx=\frac{2tant}{cos^2t}\cdot dt\)

\(\Rightarrow\int\sqrt{\frac{x+1}{x}}\cdot\frac{dx}{2}=\int\sqrt{\frac{tan^2t+1}{tan^2t}}\cdot\frac{tant}{cos^2t}dt\\ =\int\frac{1}{sin^2t}\cdot\frac{sint}{cos^3t}dt=\int\frac{d\left(cost\right)}{cos^3t\left(1-cos^2t\right)}=...\)