Tích phân [Lần 4]

Bài 1. Tính tích phân  $$I = \int_0^1 \frac{x}{1+x^4} \text{dx}$$

Ta có:
$$I = \int_0^1 \frac{x}{1+x^4} \text{dx} =\dfrac{1}{2}\int_0^1 \frac{1}{1+x^4} dx^2$$
Đặt:
$x^2=tant \Rightarrow dx^2=\dfrac{1}{cos^2t}dt$
Thì:
$$I =\dfrac{1}{2}\int_0^1\frac{1}{1+x^4} dx^2= \dfrac{1}{2}\int_0^\dfrac{\pi}{4}  \frac{1}{cos^2t(1+tan^2t)} dt=\dfrac{1}{2}\int_0^\dfrac{\pi}{4} dt=\dfrac{1}{2}t\bigg|_0^\dfrac{\pi}{4} =\dfrac{\pi}{8}$$

Bài 2. Tính tích phân $$I = \displaystyle \int_{\frac{\pi }{6}}^{\frac{\pi }{3}} \dfrac{x + \tan x}{x - \tan x} dx + \displaystyle\int_{\frac{\pi }{6}}^{\frac{\pi }{3}} \left( \dfrac{x\tan x}{x - \tan x} \right)^2dx$$

$\ (x - \tan x)'=1 - (1+\tan^2 x)=-\tan^2 x$
Vậy ta sẽ viết lại tích phân đề bài cho thành : $$I = \displaystyle \int_{\frac{\pi }{6}}^{\frac{\pi }{3}} \dfrac{x + \tan x}{x - \tan x} dx + \displaystyle\int_{\frac{\pi }{6}}^{\frac{\pi }{3}} \left( \dfrac{x\tan x}{x - \tan x} \right)^2dx =I_1 +I_2$$ Với $I_2 = \displaystyle\int_{\frac{\pi }{6}}^{\frac{\pi }{3}} \left( \dfrac{x\tan x}{x - \tan x} \right)^2dx.$Từ sự chú ý ta sẽ tính $I_2$ bằng phương pháp từng phần như sau :
Đặt $\begin{cases}u =x^2 \\ \mbox{d}v = \dfrac{\tan^2 x}{(x - \tan x)^2}\mbox{d}x = - \dfrac{\mbox{d}(x- \tan x)}{(x - \tan x)^2} \end{cases} \Rightarrow \begin{cases} \mbox{d}u =2x \mbox{d}x \\ v = \dfrac{1}{x - \tan x} \end{cases}.$ Vậy lúc đó ta có : $$I_2 = \left. \dfrac{x^2}{x - \tan x} \right|_{\frac{\pi}{6}}^{\frac{\pi}{3}} - 2\displaystyle \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \dfrac{x}{x - \tan x}\mbox{d}x$$ Tới đây các bạn hãy chú ý thêm một điều nữa đó là : $\ \dfrac{x + \tan x}{x - \tan x} = \dfrac{2x - x +\tan x }{x - \tan x}= \dfrac{2x}{x - \tan x} -1 .$ Điều này dẫn đến ta có : $$I_1 = \displaystyle \int_{\frac{\pi }{6}}^{\frac{\pi }{3}} \dfrac{x + \tan x}{x - \tan x} \mbox{d}x=   2 \displaystyle \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \dfrac{\tan x}{x - \tan x}\mbox{d}x - \displaystyle \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \mbox{d}x=2 \displaystyle \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \dfrac{\tan x}{x - \tan x}\mbox{d}x - \left. x \right|_{\frac{\pi}{6}}^{\frac{\pi}{3}}$$ Từ đây ta có $I =I_1+I_2 = \left. \left(\dfrac{x^2}{x - \tan x} -x \right) \right|_{\frac{\pi}{6}}^{\frac{\pi}{3}}= \dfrac{\pi^2}{3} \left( \dfrac{1}{\pi -3 \sqrt 3} - \dfrac{1}{2(\pi -2 \sqrt 3)} \right) - \dfrac{\pi}{6}$

Bài 3. Tính tích phân $$I = \int\limits_0^2 {\dfrac{{x + 2}}{{(x + 1)({x^2} + 2x + 4)}}dx}$$

Ta có:
$$I = \int\limits_0^2 {\dfrac{{x + 2}}{{(x + 1)({x^2} + 2x + 4)}}dx} = \int\limits_0^2 {\dfrac{{1}}{{{x^2} + 2x + 4}}+\dfrac{1}{(x+1)({x^2} + 2x + 4)}dx}$$
Mà:
$$I_{1}=\int\limits_0^2 \dfrac{1}{{x^2} + 2x + 4}dx=\int\limits_0^2 \dfrac{1}{(x+1)^2+3}dx=\int\limits_0^2 \dfrac{1}{(x+1)^2+3}d(x+1)$$
$$I_{2}=\int\limits_0^2 \dfrac{1}{(x+1)(x^2 + 2x + 4)}dx=\int\limits_0^2 \dfrac{1}{(x+1)[(x+1)^2 +3]}d(x+1)=\dfrac{1}{2}\int\limits_0^2 \dfrac{1}{(x+1)^2[(x+1)^2 +3]}d(x+1)^2$$

Bài 4. Tính tích phân: $$I=\displaystyle \int_0^4 \dfrac{x+1}{(1+\sqrt{1+2x})^2}dx$$

Đặt:
$t=\sqrt{1+2x} \Rightarrow x=\dfrac{t^2-1}{2}, dx=t.dt$
Vậy:
$$I=\displaystyle \int_0^4 \dfrac{x+1}{(1+\sqrt{1+2x})^2}dx=\displaystyle \int_1^3 \dfrac{(\dfrac{t^2-1}{2}+1)t}{(1+t)^2}dt=\dfrac{1}{2}\displaystyle \int_1^3 \dfrac{(t^2+1)t}{(1+t)^2}dt=\dfrac{1}{2} \displaystyle \int_1^3 \dfrac{(t+1)^2.t-2t^2}{(1+t)^2}dt$$ $$=\dfrac{1}{2} \displaystyle \int_1^3tdt-\dfrac{1}{2} \displaystyle \int_1^3 \dfrac{2t^2}{(1+t)^2}dt$$
Mà:
$$I_{1}=\dfrac{1}{2} \displaystyle \int_1^3tdt=\dfrac{t^2}{4}\bigg|^3_1=2.$$
$$I_{2}=\dfrac{1}{2} \displaystyle \int_1^3 \dfrac{2t^2}{(1+t)^2}dt.=\displaystyle \int_1^3 \dfrac{(t+1)^2-2t-1}{(1+t)^2}dt.=\displaystyle \int_1^3 dt-\displaystyle \int_1^3 \dfrac{2t+1}{(1+t)^2}dt.$$ $$=\displaystyle \int_1^3dt-\displaystyle \int_1^3 \dfrac{2}{1+t}dt+\displaystyle \int_1^3 \dfrac{1}{(1+t)^2}dt..=\left(t-2.\ln|t+1|-\dfrac{1}{1+t}\right)\bigg|^3_1=\dfrac{9}{4}-\ln4.$$
Vậy:
$$I=-\dfrac{1}{4}+\ln4.$$

Bài 5. Tính tích phân: $$I=\displaystyle \int_1^{\sqrt[3]{2}}\sqrt{x-\dfrac{1}{x^2}}dx$$

$I=\displaystyle \int_1^{\sqrt[3]{2}}\sqrt{x-\dfrac{1}{x^2}}dx=\displaystyle \int_1^{\sqrt[3]{2}}x^{-1}(-1+x^3)^{\frac{1}{2}}dx$
Đặt: $t=(-1+x^3)^{\frac{1}{2}}\Longleftrightarrow t^2=x^3-1\Longrightarrow 2tdt=3x^2dx\Longrightarrow dx=\dfrac{2tdt}{3x^2}=\dfrac{2tdt}{3\sqrt[3]{(t^2+1)^2}}$
Đổi cận: $t_{(1)}=0;t_{(\sqrt[3]2)}=1$
Do đó:
$\begin{aligned}I=\displaystyle \int\limits_0^1\dfrac{2t^2dt}{3\sqrt[3]{t^2+1}\sqrt[3]{(t^2+1)^2}}&=\dfrac{2}{3}\int\limits_0^1\dfrac{t^2}{t^2+1}dt\\&=\dfrac{2}{3}\int\limits_0^1\dfrac{t^2+1}{t^2+1}dt-\dfrac{2}{3}\int\limits_0^1\dfrac{1}{t^2+1}dt\\&= \dfrac{2}{3}\int\limits_0^1dt-\dfrac{2}{3}I'\end{aligned}$
Để tính I' ta chỉ việc đặt $t=\tan u$