Tích phân

3/ $P = \int\limits_1^e {\frac{{1 + {x^2}lnx}}{{x + {x^2}lnx}}} dx$

$= \int\limits_1^e {\frac{{x + {x^2}lnx - x + 1}}{{x + {x^2}lnx}}} dx$

$= \int\limits_1^e {\left( {1 - \frac{{x - 1}}{{x + {x^2}lnx}}} \right)} dx$

$= \int\limits_1^e {\left( {1 - \frac{{x + x\ln x - 1 - x\ln x}}{{x(1 + xlnx)}}} \right)} dx$

$= \int\limits_1^e {\left( {1 - \frac{{1 + \ln x}}{{1 + xlnx}} + \frac{1}{x}} \right)} dx$

$= e + 1 - \int\limits_1^e {\frac{{1 + \ln x}}{{1 + xlnx}}dx}$

Tới đây đặt $t = 1 + x\ln x$ là xong.

2/ $I = \int\limits_0^{\pi /2} {\frac{{{{\cos }^2}x}}{{3\sin x + 4\cos x}}} dx$

Đặt $J = \int\limits_0^{\pi /2} {\frac{{{{\sin }^2}x}}{{3\sin x + 4\cos x}}} dx$

* $16I - 9J = \int\limits_0^{\pi /2} {\frac{{16{{\cos }^2}x - 9{{\sin }^2}x}}{{3\sin x + 4\cos x}}} dx = \int\limits_0^{\pi /2} {(4\cos x - 3\sin x)} dx = 1  (1)$.

* $I + J = \int\limits_0^{\pi /2} {\frac{1}{{3\sin x + 4\cos x}}} dx$

$3\sin x + 4\cos x = 6\sin \frac{x}{2}\cos \frac{x}{2} + 4{\cos ^2}\frac{x}{2} - 4{\sin ^2}\frac{x}{2}$

$=  - 2{\cos ^2}\frac{x}{2}\left( {2{{\tan }^2}\frac{x}{2} - 3\tan \frac{x}{2} - 2} \right)$

Đặt $t = \tan \frac{x}{2} \Rightarrow dt = \frac{{dx}}{{2{{\cos }^2}\frac{x}{2}}}$

$I + J =  - \int\limits_0^{\pi /2} {\frac{{dx}}{{2{{\cos }^2}\frac{x}{2}\left( {2{{\tan }^2}\frac{x}{2} - 3\tan \frac{x}{2} - 2} \right)}}}  =  - \int\limits_0^1 {\frac{{dt}}{{2{t^2} - 3t - 2}}}$

$=  - \frac{1}{5}\int\limits_0^1 {\left( {\frac{1}{{t - 2}} - \frac{2}{{2t + 1}}} \right)dt = \frac{1}{5}} \ln 6   (2).$

* Từ (1) và (2) suy ra $I = \frac{{5 + 9\ln 6}}{{125}}$