# Bài 2: Tích phân

26 tháng 8 2020 lúc 13:59

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

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

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

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

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

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26 tháng 8 2020 lúc 14:05

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 $x+1=2sinu\Rightarrow dx=2cosu.du$

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

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

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26 tháng 8 2020 lúc 14:38

c/

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

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

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

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

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

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

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