[tex]I= \int\limits^3_1 {\frac{1}{\sqrt{x}(x+3)}} \, dx \\\\y=\sqrt{\frac{x}{3}},\ \ dy=\frac{1}{2\sqrt{3x}}dx,\ dx=2\sqrt{3}*\sqrt{3}\ y\ dy\\\\ I= \int\limits^{1}_{\frac{1}{\sqrt{3}}} {\frac{6y}{\sqrt{3}y(3y^2+3)}} \, dy \\\\=\frac{2}{\sqrt{3}}[tan^{-1}y]_{\frac{1}{\sqrt{3}}}^1\\\\=\frac{2}{\sqrt{3}}[\frac{\pi}{4}-\frac{\pi}{6}]=\frac{\pi}{6\sqrt{3}}\\[/tex]
[tex]I= \int\limits^3_2 {\frac{1}{\sqrt{4x-x^2}}} \, dx\\\\I=\int\limits^3_2 {\frac{1}{\sqrt{+4-4+4x-x^2}}} \, dx \\\\I=\int\limits^3_2 {\frac{1}{\sqrt{4-(x-2)^2}}} \, dx \\\\y=\frac{x-2}{2},\ \ dy=dx/2\\\\I= \int\limits^{\frac{1}{2}}_0 {\frac{1}{2\sqrt{1-y^2}}} \, 2*dy\\\\=[Sin^{-1}y]_0^{\frac{1}{2}}=\frac{\pi}{6}\\[/tex]
[tex]I= \int\limits^1_0 {\frac{x^4(1-x)^4}{1+x^2}} \, dx\\\\\frac{x^4(1-x)^4}{1+x^2}=\frac{x^4(x^4-4x^3+6x^2-4x+1)}{1+x^2}\\\\=x^6-4x^5+5x^4-4x^2+4-\frac{4}{1+x^2}\\\\I= \int\limits^1_0 { x^6-4x^5+5x^4-4x^2+4-\frac{4}{1+x^2}} \, dx\\\\=[\frac{x^7}{7}-\frac{4x^6}{6}+\frac{5x^5}{5}-\frac{4x^3}{3}+4x-4*tan^{-1}x ]^1_0\\\\=\frac{1}{7}-\frac{2}{3}+1-\frac{4}{3}+4-\frac{4*\pi}{4}\\\\=\frac{22}{7}-\pi\\[/tex]
on fait la division de l'expression polynome en x: (x⁸-4x⁷+6x⁶-4x⁵+x⁴) par (1+x²)
1+x² ) x⁸ - 4 x⁷ + 6 x⁶ - 4 x⁵ + x⁴ ( x⁶ - 4 x⁵ + 5x⁴ - 4x² + 4
x⁸ + x⁶
==================
- 4 x⁷ + 5 x⁶ - 4 x⁵
- 4 x⁷ - 4 x⁵
====================
5 x⁶ + x⁴
5 x⁶ + 5 x⁴
====================
- 4 x⁴
- 4 x⁴ - 4x²
===============
4 x²
4 x² + 4
=================
-4
