Tính tích phân

N

nguyenbahiep1

câu 1

[TEX]\int\limits_{0}^{1}ln(1+x^2)dx[/TEX]

[laTEX]u = ln(1+x^2) \Rightarrow du = \frac{2x}{x^2+1} \\ \\dv = dx \Rightarrow v = x \\ \\ I = xln(1+x^2) \big|_0^1 - \int_{0}^{1}\frac{2x^2}{x^2+1} = ln 2 - I_1 \\ \\ I_1 = 2\int_{0}^{1}(1 - \frac{1}{x^2+1})dx = 2(x - arctanx) \big|_0^1 = 2(1 - \frac{\pi}{4}) \\ \\ I = ln2 - 2(1 - \frac{\pi}{4})[/laTEX]
 
N

nguyenbahiep1

3. [TEX]\int\limits_{2}^{3}ln(x^2-x)dx[/TEX]

tích phân từng phần

[laTEX]u = ln(x^2-x) \\ \\ dv = dx [/laTEX]
 
N

nguyenbahiep1

4. [TEX]\int\limits_{0}^{\frac{\pi}{2}}\frac{(sin2x+sinx)dx}{\sqrt{1+3cosx}}[/I] [laTEX]\frac{(sin2x+sinx)dx}{\sqrt{1+3cosx}} = \frac{sinx(2cosx+1)dx}{\sqrt{1+3cosx}} \\ \\ \sqrt{1+3cosx} = u [/laTEX][/TEX]
 
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