Tích phân [Lần 3]

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

Đặt $t=e^x . x$, ta có: $\mathrm{dt = (e^x + e^x x) dx}$.
[LIST]
[*]$ x=1 \Rightarrow t=e$
[*]$x=2 \Rightarrow t = 2e^2$
[/LIST]
$\displaystyle I = \int_{e}^{2e^2} \frac{\mathrm{dt}}{t(t+1)} = (\ln \text{|}t\text{|} - \ln \text{|} t+1 \text{|} ) \left| \begin{aligned} &2e^2 \\ \\ &e \end{aligned} \right. = \ln {\Bigg| \frac{\frac{2e^2}{2e^2+1}}{\frac{e}{e+1}} }\Bigg| = \ln \Bigg| \frac{2e(e+1)}{2e^2+1}\Bigg|$

Bài 2. Tính tích phân: $$I= \int_1^2 \frac{1-x^{2012}}{x(1+x^{2012})} \text{dx}$$

Ta có:
$$I= \int_1^2 \frac{1-x^{2012}}{x(1+x^{2012})} \text{dx}=\int_1^2 \frac{1+x^{2012}-2x^{2012}}{x(1+x^{2012})} \text{dx}=\int_1^2 \frac{1}{x} \text{dx}-\int_1^2 \frac{2x^{2011}}{1+x^{2012}} \text{dx}$$
Mà:
$$\int_1^2 \frac{2x^{2011}}{1+x^{2012}} \text{dx}=\dfrac{1}{1006}\int_1^2 \frac{1}{1+x^{2012}} d(1+x^{2012})$$
Vậy:
$$I=\left(\ln \bigg| x \bigg|-\dfrac{\ln \bigg|1+x^{2012} \bigg|}{1006}\right)\bigg|^2_1$$

Bài 3. Tính tích phân: $$\mathrm{ I=\int_2^3 \frac{x^4}{(x^2-1)^2} dx} $$

Ta có: $$\mathrm{ I=\int \limits_2^3 \frac{\left(x^2-1+1\right)^2}{(x^2-1)^2} dx=\int \limits_2^3 \left(1+\dfrac{2}{x^2-1}+\dfrac{1}{(x^2-1)^2}\right)dx=x\bigg|_2^3 +2\int \limits_2^3 \dfrac{1}{x^2-1}dx+\int \limits_2^3 \dfrac{dx}{(x^2-1)^2}}$$
Tính: $$\mathrm{ I_1=\int \limits_2^3 \dfrac{dx}{x^2-1}=\dfrac{1}{2}\int \limits_2^3 \left(\dfrac{1}{x-1}-\dfrac{1}{x+1}\right)dx=\dfrac{1}{2}\ln \left|\dfrac{x-1}{x+1}\right|\bigg|_2^3=\dfrac{1}{2}\ln \dfrac{3}{2}}$$
$$\mathrm{ \begin{aligned} I_2&=\int \limits_2^3 \dfrac{dx}{(x^2-1)^2}=\dfrac{1}{4}\int \limits_2^3 \left(\dfrac{1}{(x-1)^2}+\dfrac{1}{(x+1)^2}-\dfrac{2}{x^2-1}\right)dx\\
&=-\dfrac{1}{4}\left(\dfrac{1}{x-1}+\dfrac{1}{x+1}\right)\bigg|_2^3-\dfrac{1}{2}I_1=\dfrac{7}{48}-\dfrac{1}{4}\ln \dfrac{3}{2}\end{aligned}}$$
Vậy: $I=x\bigg|_2^3+2I_1+I_2=\dfrac{55}{48}+\dfrac{3}{4}\ln \dfrac{3}{2}$

Bài 4. Tính tích phân $$\mathrm{I = \int_0^1 \frac{x^{2011}}{(1+x^2)^{1007}} dx}$$

$\mbox{I} = \dfrac12.\displaystyle\int_0^1\left(\dfrac{x^2}{x^2+1}\right)^{1005}.\dfrac{2x}{(x^2+1)^2}\mbox{d}x = \dfrac12.\displaystyle\int_0^1\left(\dfrac{x^2}{x^2+1}\right)^{1005}.\mbox{d}{\left(\dfrac{x^2}{x^2+1} \right)} = \dfrac1{2012}.\left(\dfrac{x^2}{x^2+1} \right) \bigg|^1_0 =\dfrac1{4024} $

Bài 5. Tính tích phân $$\mathrm{I= \int_1^2 \frac{x^2-3}{x(x^4+3x^2+2)} dx}$$

Ta có:
$$\mathrm{I= \int_1^2 \frac{x^2-3}{x(x^4+3x^2+2)} dx}=\int_1^2 \frac{2x^2-(x^2+1)-(x^2+2)}{x(x^2+1)(x^2+2)}dx$$
$$=\int_1^2 \frac{3x}{(x^2+1)(x^2+2)} dx -\int_1^2 \frac{1}{x(x^2+2)} dx-\int_1^2 \frac{1}{x(x^2+1)} dx$$
Mà:
$$I_{1}=\int_1^2 \frac{3x}{(x^2+1)(x^2+2)} dx =\dfrac{3}{2}\int_1^2 \dfrac{1}{x^2+1}-\dfrac{1}{x^2+2} dx^2=\dfrac{3}{2}\left(\ln\bigg|x^2+1\bigg|- \ln\bigg|x^2+2\bigg|\right)\bigg|^2_1$$
$$I_{2}=\int_1^2 \frac{1}{x(x^2+2)} dx=\dfrac{1}{4}\int_1^2 \dfrac{1}{x^2}-\dfrac{1}{x^2+2} dx^2=\dfrac{1}{4}\left(\ln\bigg|x^2\bigg|- \ln\bigg|x^2+2\bigg|\right)\bigg|^2_1$$
$$I_{2}=\int_1^2 \frac{1}{x(x^2+1)} dx=\dfrac{1}{2}\int_1^2 \dfrac{1}{x^2}-\dfrac{1}{x^2+2} dx^2=\dfrac{1}{2}\left(\ln\bigg|x^2\bigg|- \ln\bigg|x^2+1\bigg|\right)\bigg|^2_1$$