viết lại hẳn hoi đi chứ ,cứ viết kiểu z có ma trả lời 

cẩu thả

Chế đúng ko

ko có kết quả linh tinh

bít oóc rồi hả

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

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

a.

\(I=\int\frac{\frac{1}{2}\left(2x-2\right)+7}{\sqrt{x^2-2x+10}}dx=\frac{1}{2}\int\frac{2x-2}{\sqrt{x^2-2x+10}}dx+7\int\frac{1}{\sqrt{x^2-2x+10}}dx=\frac{1}{2}I_1+7I_2\)

Xét \(I_1=\int\frac{2x-2}{\sqrt{x^2-2x+10}}dx=\int\frac{d\left(x^2-2x+10\right)}{\sqrt{x^2-2x+10}}=2\sqrt{x^2-2x+10}+C_1\)

Xét \(I_2=\int\frac{dx}{\sqrt{x^2-2x+10}}=\int\frac{dx}{\sqrt{\left(x-1\right)^2+9}}\)

Đặt

\(u=x-1+\sqrt{\left(x-1\right)^2+10}\Rightarrow du=\left(1+\frac{\left(x-1\right)}{\sqrt{\left(x-1\right)^2+10}}\right)dx=\frac{x-1+\sqrt{\left(x-1\right)^2+10}}{\sqrt{\left(x-1\right)^2+10}}dx\)

\(\Rightarrow du=\frac{u}{\sqrt{\left(x-1\right)^2+10}}dx\Rightarrow\frac{dx}{\sqrt{\left(x-1\right)^2+10}}=\frac{du}{u}\)

\(\Rightarrow I_2=\int\frac{du}{u}=ln\left|u\right|+C_2=ln\left|x-1+\sqrt{x^2-2x+10}\right|+C_2\)

\(\Rightarrow I=\sqrt{x^2-2x+10}+7ln\left|x-1+\sqrt{x^2-2x+10}\right|+C\)

2.

\(I=\int\frac{\frac{1}{2}\left(2x+2\right)-1}{\sqrt{3-2x-x^2}}dx=\frac{1}{2}\int\frac{2x+2}{\sqrt{3-2x-x^2}}dx-\int\frac{1}{\sqrt{3-2x-x^2}}dx=\frac{1}{2}I_1-I_2\)

Xét \(I_1=\int\frac{2x+2}{\sqrt{3-2x-x^2}}dx=-\int\frac{d\left(3-2x-x^2\right)}{\sqrt{3-2x-x^2}}=-2\sqrt{3-2x-x^2}+C_1\)

Xét \(I_2=\int\frac{1}{\sqrt{3-2x-x^2}}dx=\int\frac{1}{\sqrt{4-\left(x+1\right)^2}}dx\)

Đặt \(x+1=2sinu\Rightarrow dx=2cosu.du\)

\(\Rightarrow I_2=\int\frac{2cosu.du}{2.cosu}=\int du=u+C_2=arcsin\left(\frac{x+1}{2}\right)+C_2\)

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

c/

\(I=\int\frac{1-\sqrt{x}}{\sqrt{1-x}}dx\)

Đặt \(\sqrt{x}=sint\Rightarrow x=sin^2t\Rightarrow dx=2sint.cost.dt\)

\(\Rightarrow I=\int\frac{2sint.cost\left(1-sint\right)}{\sqrt{1-sin^2t}}dt=\int\frac{2sint.cost\left(1-sint\right)}{cost}dt=\int\left(2sint-2sin^2t\right)dt\)

\(=\int\left(2sint+cos2t-1\right)dt=-2cost+\frac{1}{2}sin2t-t+C\)

\(=-2\sqrt{1-sin^2t}+\frac{1}{2}sint\sqrt{1-sin^2t}-t+C\)

\(=-2\sqrt{1-x}+\frac{1}{2}\sqrt{x\left(1-x\right)}-arcsin\left(\sqrt{x}\right)+C\)

lm jup mk di m.n

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

3) Đặt \(t=\sqrt{1+\sqrt[3]{x^2}}\Rightarrow t^2-1=\sqrt[3]{x^2}\Leftrightarrow x^2=\left(t^2-1\right)^3\)

\(d\left(x^2\right)=d\left[\left(t^2-1\right)^3\right]\Leftrightarrow2x\cdot dx=6t\left(t^2-1\right)^2\cdot dt\)

\(I_3=\int\frac{3t\left(t^2-1\right)^2}{t}dt=3\int\left(t^4-2t^2+1\right)dt=...\)

5) Đặt \(\frac{2+x}{2-x}=4t^3\Leftrightarrow4t^3=\frac{4}{2-x}-1\)

\(d\left(4t^3\right)=d\left(\frac{4}{2-x}-1\right)\Leftrightarrow3t^2dt=\frac{1}{\left(2-x\right)^2}dx\)

\(I_5=\int\frac{3t^2}{t\sqrt[3]{4}}dt=\frac{3}{\sqrt[3]{4}}\int tdt=...\)