Toán 12 [Đấu trường toán 12_ver2] Tích phân

[trantien.hocmai]

$$I=\int_1^e \dfrac{2x^2+x(1+2\ln x)+\ln ^2x}{(x^2+x\ln x)^2}dx=\int_1^e \dfrac{x^2+x+(x+\ln x)^2}{x^2(x+\ln x)^2}dx \\
=\int_1^e \dfrac{x^2+x}{x^2(x+\ln x)^2}dx+\int_1^e \dfrac{dx}{x^2}=I_1+I_2 \\
I_1=\int_1^e \dfrac{x^2+x}{x^2(x+\ln x)^2}dx=\int_1^e \dfrac{1+\dfrac{1}{x}}{(x+\ln x)^2}dx \\$$
$\text{đặt }u=x+\ln x \rightarrow du=(1+\dfrac{1}{x})dx \\
\text{đổi cận }x=1 \rightarrow u=1, x=e \rightarrow u=1+e \\$
$$\rightarrow I_1=\int_1^{1+e} \dfrac{du}{u^2}=-\dfrac{1}{u}|_1^{1+e}= \\$$
$$I_2=\int_1^e \dfrac{dx}{x^2}=-\dfrac{1}{x}|_1^e= \\$$
$$I=\int_0^{\dfrac{\pi}{2}} \dfrac{x^2+\sin ^2x-\sin x}{x+\cos x}dx=\int_0^{\dfrac{\pi}{2}} \dfrac{x^2-\cos ^2x+\sin ^2x+\cos ^2x-\sin x}{x+\cos x}dx \\ =\int_0^{\dfrac{\pi}{2}} \dfrac{(x-\cos x)(x+\cos x)}{x+\cos x}dx+\int_0^{\dfrac{\pi}{2}} \dfrac{1-\sin x}{x+\cos x}dx=I_1+I_2 \\
I_1=\int_0^{\dfrac{\pi}{2}} \dfrac{(x-\cos x)(x+\cos x)}{x+\cos x}dx=\int_0^{\dfrac{\pi}{2}} (x-\cos x)=(\dfrac{1}{2}x^2-\sin x)|_0^{\dfrac{\pi}{2}} \\$$
$$I_2=\int_0^{\dfrac{\pi}{2}} \dfrac{1-\sin x}{x+\cos x}dx=\int_0^{\dfrac{\pi}{2}} \dfrac{d(x+\cos x)}{x+\cos x}=\ln|x+\cos x||_0^{\dfrac{\pi}{2}} \\$$
$$$$

 

[trantien.hocmai]

$\text{đề nữa nè} \\
I=\int_{-\dfrac{\pi}{4}}^{\dfrac{\pi}{4}} \dfrac{x^7-x^5+x^3-x+1}{\cos ^4x}dx \\
I=\int_{-1}^1 \ln (x+\sqrt{x^2+1})dx \\
I=\int_{-2}^2 [\ln (x+\sqrt{1+x^2})]^5 dx$

 

[quynhsieunhan]

$I = \int\limits_{-1}^{1}ln(x + \sqrt{x^2 + 1})dx$
Đặt $\left\{ \begin{array}{l} ln(x + \sqrt{x^2 + 1}) = u \\ dx = dv \end{array} \right.$ \Rightarrow $\left\{ \begin{array}{l} du = \frac{dx}{\sqrt{x^2 + 1}} \\ v = x \end{array} \right.$
\Rightarrow $I = xln(x + \sqrt{x^2 + 1})|_{-1}^{1} - \int\limits_{-1}^{1}\frac{xdx}{\sqrt{x^2 + 1}}$
= $ln(1 + \sqrt{2}) + ln(\sqrt{2} - 1) - \sqrt{x^2 + 1}|_{-1}^{1}$
= 0

 

[quynhsieunhan]

Đây nữa nhé
tính nguyên hàm
$I = \int\frac{x(3cot^22x - cos^2x) + sinx(cosx - xsinx)}{2cos4x + 1}dx$
$I = \int\limits_{0}^{ln2}\frac{e^xdx}{(e^x - 9)\sqrt{3e^x - 2}}$

 

[vivietnam]

Thử ý này nha: (nguyên hàm thôi )
$\int\frac{sin^4x}{sin^4x + cos^4x}dx$

đặt
$I=\int\dfrac{sin^4x}{sin^4x + cos^4x}dx$
$J=\int\dfrac{cos^4x}{sin^4x + cos^4x}dx$
lập hệ phương trình
$I+J=\int \dfrac{sin^4x+cos^4x}{sin^4x+cos^4x}dx=\int dx=x $
$I-J=\int \dfrac{sin^4x-cos^4x}{sin^4x+cos^4x}dx=\int \dfrac{sin^2x-cos^2x}{1-2sin^2x.cos^2x}dx=\int \dfrac{-cos2xdx}{1-\dfrac{sin^22x}{2}}$
$I-J=\int \dfrac{-d(sin2x)}{2-sin^22x}=.......$
biết tổng hiệu I,J tính được I

 

[thuyanh_tls1417]

From: https://www.facebook.com/kolkatamaths​

Solution