Số lớn gấp 10 làn số bé và thêm 1 đơn vị

Số bé là:(2870-1):(1+10)=........

Số lớn là:2870-...=.....

Đề sai nhé

1) Đặt \(2+lnx=t\Leftrightarrow x=e^{t-2}\Rightarrow dx=e^{t-2}dt\)

\(I_1=\int\left(\frac{t-2}{t}\right)^2\cdot e^{t-2}\cdot dt=\int\left(1-\frac{4}{t}+\frac{4}{t^2}\right)e^{t-2}dt\\ =\int e^{t-2}dt-4\int\frac{e^{t-2}}{t}dt+4\int\frac{e^{t-2}}{t^2}dt\)

Có:

\(4\int\frac{e^{t-2}}{t^2}dt=-4\int e^{t-2}\cdot d\left(\frac{1}{t}\right)=-\frac{4\cdot e^{t-2}}{t}+4\int\frac{e^{t-2}}{t}dt\\ \Leftrightarrow4\int\frac{e^{t-2}}{t^2}dt-4\int\frac{e^{t-2}}{t^{ }}dt=-\frac{4\cdot e^{t-2}}{t}\)

Vậy \(I_1=\int e^{t-2}dt-\frac{4\cdot e^{t-2}}{t}=e^{t-2}-\frac{4e^{t-2}}{t}+C\)

1)

\(\int\frac{tan^3x}{cos2x}dx=\int\frac{sin^3x}{cos^3x\cdot\left(2cos^2x-1\right)}dx=\int\frac{1-cos^2x}{cos^3x\left(2cos^2x-1\right)}\cdot sinx\cdot dx\\ =\int\frac{1-cos^2x}{cos^3x\left(2cos^2x-1\right)}d\left(cosx\right)=...\)

Đặt x=2t, dx=2dt

\(2sinx+5cosx+3=2sin2t+5cos2t+3\\ =4sint\cdot cost+5\left(cos^2t-sin^2t\right)+3\left(sin^2t+cos^2t\right)\\ =-2sin^2t+4sint\cdot cost+8cos^2t\)

Ta có:

\(I=\int\frac{2dt}{-2sin^2t+4sint\cdot cost+8cos^2t}\\ =\int\frac{\frac{dt}{cos^2t}}{-tan^2t+2tant+4}=\int\frac{d\left(tant\right)}{-tan^2t+2tant+4}\\ =\int\frac{-d\left(tant\right)}{\left(tant-1+\sqrt{5}\right)\left(tant-1-\sqrt{5}\right)}\\ =\frac{1}{2\sqrt{5}}\int\left(\frac{1}{tant-1+\sqrt{5}}-\frac{1}{tant-1-\sqrt{5}}\right)dt\)

\(=\frac{1}{2\sqrt{5}}ln\left|\frac{tant-1+\sqrt{5}}{tant-1+\sqrt{5}}\right|+C\)

....

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2, Mr. Brown ofent gives advice to his students

3, The girls have to learn how to make cakes

4, She is at school now and she is teaching English at my class

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đặt \(x=\frac{\sqrt{3}}{cost};\forall t\in\left(0;\frac{\pi}{2}\right)\Rightarrow tant>0\)

\(dx=d\left(\frac{\sqrt{3}}{cost}\right)=\frac{-\sqrt{3}sint}{cos^2t}dt\)

Thay vào, ta có \(\int\frac{\sqrt{3}\cdot\frac{-\sqrt{3}sint}{cos^2t}}{\frac{\sqrt{3}}{cost}\sqrt{\frac{3}{cos^2t}-3}}dt=\int\frac{-3\cdot\frac{sint}{cos^2t}}{\frac{3}{cost}\cdot\sqrt{tan^2t}}dt=\int\frac{-sint}{cost\cdot tant}dt=-\int dt=-t+C\)

Bây giờ thay t vào là ra

\(\int\frac{dx}{x^5\left(x^4+5\right)}=\frac{1}{25}\int\left(\frac{5}{x^5}-\frac{1}{x}+\frac{x^3}{x^4+5}\right)dx\)

\(\int\frac{5}{x^5}dx=-\frac{5}{4}.x^{-4}+C\)

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

\(\int\frac{x^3}{x^4+5}dx=\frac{1}{4}\cdot\int\frac{d\left(x^4+5\right)}{x^4+5}=\frac{ln\left(x^4+5\right)}{4}+C\)

5) Đặt \(x=3sint\Leftrightarrow dx=3cost\cdot dt\)

\(I_5=\int\frac{3cost}{\sqrt{9-9sin^2t}}dt=\int\frac{3cost}{3cost}dt=\int dt=t+C\)

.....

4) \(I_3\int\frac{dx}{x\left(x-1\right)\left(3x-4\right)}=\int\left(\frac{A}{x}+\frac{B}{x-1}+\frac{C}{3x-4}\right)dx=...\)

Bạn tự tìm A,B,C nhé

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

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