Tích phân [Lần 7]

Bài 1.  Tính tích phân: $$I=\displaystyle \int_1^e \dfrac{x^2.\ln x+1}{x(1+x.\ln x)}dx$$

$I = \int_1^e \frac{x ( x \ln x + 1) + ( 1- x) }{x ( 1 + x \ln x )} dx $

$= \int_1^e dx + \int_1^e \frac{\frac{1}{x^2} - \frac{1}{x}}{ \frac{1}{x} + \ln x} dx $

$= (e-1) - \int_1^e \frac{d( \ln x + \frac{1}{x})}{ \frac{1}{x} + \ln x}$
$= e-1 - \ln \big| \frac{1}{x} + \ln x \big| \bigg|_1^e = e - \ln (e+1)$

Bài 2. Tính tích phân: $$I=\displaystyle \int^{\frac{\pi}{2}}_{\frac{\pi}{6}}. \sin x \sqrt{\sin^2x+\dfrac{1}{2}}dx$$

$$I = - \int^{\frac{\pi}{2}}_{\frac{\pi}{6}} \sqrt{\dfrac32- \cos^2 x} d( \cos x) $$
Đặt $\cos x =\sqrt{ \dfrac32} \sin t \Rightarrow d (\cos x) = \sqrt{\dfrac32} \cos t  dt $.
Đổi cận: $ x = \frac{\pi}{6} \Rightarrow t = \frac{\pi}{4},\ x = \frac{\pi}{2} \Rightarrow t = 0 $
$$I =  \int_0^{\frac{\pi}{4}} \dfrac32 . \cos^2 t dt $$

Bài 3. Tính tích phân: $$I=\displaystyle \int_5^7 \dfrac{1+\ln^2x }{\sqrt{x.\ln^3x}}dx$$

$$\begin{aligned} I &= \int_5^7 \dfrac{\ln^2 x + \ln x + 1 - \ln x }{\ln x \sqrt{x \ln x}} dx \\ & = \int_5^7 \dfrac{\ln x + 1 }{\sqrt{x \ln x}} - \dfrac{\ln x - 1}{\ln^2 x \sqrt{ \frac{x}{\ln x}}} dx \\ & = \int_5^7 \dfrac{d( x \ln x)}{\sqrt{x \ln x} } - \int_5^7 \dfrac{d( \frac{x}{\ln x} )}{\sqrt{ \frac{x}{\ln x} } }   \\ & = \left( 2( \sqrt{x \ln x} - \sqrt{\frac{x}{\ln x}}) \right) \bigg|_5^7 \end{aligned}$$

Bài 4. Tính tích phân $\displaystyle \int_0^1 xln(x + 1)dx$

$$I = \frac12 \int_0^1 \ln (x+1) d(x^2-1) = \frac12 \left( (x^2-1) \ln (x+1) \bigg|_0^1 - \int_0^1 (x^2-1) \frac{1}{x+1} dx \right) = -\frac12 ( \frac12 x^2 - x) \bigg|_0^1 = \frac14$$

Bài 5. Tính $$I = \displaystyle \int_{-1}^{1}\frac{dx}{1 + x + x^2 + \sqrt{x^4 + 3x^2 + 1}}$$

Đặt $x=-t,$ ta có $\text{d}x=-\text{d}t.$ Các cận của tích phân sẽ thay đổi như sau:
[LIST]
[*]Với $x=-1$ thì $t=1.$
[*]Với $x=1$ thì $t=-1.$
[/LIST]
Như vậy, $$\begin{aligned} I&= \int\limits_{1}^{-1} \frac{-\text{d}t}{t^2-t+1+\sqrt{t^4+3t^2+1}}= \int\limits_{-1}^1 \frac{\text{d}t}{t^2-t+1+\sqrt{t^4+3t^2+1}} \\ &=\int\limits_{-1}^1 \frac{\text{d}x}{x^2-x+1+\sqrt{x^4+3x^2+1}}.\end{aligned}$$ Từ đây, ta có $$\begin{aligned} 2I&= \int\limits_{-1}^1 \left(\frac{1}{x^2+x+1+\sqrt{x^4+3x^2+1}}+\frac{1}{x^2-x+1+\sqrt{x^4+3x^2+1}}\right) \text{d} x \\ &=\int\limits_{-1}^1 \frac{2\left( x^2+1+\sqrt{x^4+3x^2+1}\right)}{\left( x^2+x+1+\sqrt{x^4+3x^2+1}\right) \left( x^2-x+1+\sqrt{x^4+3x^2+1}\right)} \text{d} x \\ &=\int\limits_{-1}^1 \frac{2\left(x^2+1+\sqrt{x^4+3x^2+1}\right)}{ \left( x^2+1+\sqrt{x^4+3x^2+1}\right)^2-x^2} \text{d} x =\int\limits_{-1}^1 \frac{\text{d}x}{x^2+1} \\ &= \int\limits_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{\text{d}(\tan x)}{\tan^2 x+1}=\int\limits_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{(\tan^2 x+1)\text{d} x}{\tan^2 x+1} =\int\limits_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \text{d} x =\left. x \right|_{-\frac{\pi}{4}}^{\frac{\pi}{4}} =\frac{\pi}{2}. \end{aligned}$$ Và ta tính được $I =\frac{\pi}{4}.$$