Tích phân
[TEX]I=\int_{1}^{2}(x^2 +x).ln(1+x^2)dx[/TEX]
[TEX]I= \int_{1}^{2}x^2.ln(1+x^2)dx + \int_{1}^{2}xln(1+x^2)dx[/TEX]
[TEX]I= A +B (1)[/TEX]
[TEX]A=\int_{1}^{2}(x^2 +x).ln(1+x^2)dx[/TEX]
Đặt
[TEX]u=ln(1+x^2) => du= \frac{2x}{1+x^2}dx[/TEX]
[TEX]dv=x^2dx => v= \frac{x^3}{3} [/TEX]
[TEX]A= \frac{8}{3}ln5 - \frac{1}{3}ln2 + \frac{2}{3}\int_{1}^{2}(\frac{x^4}{x^2+1})dx[/TEX]
[TEX]A=\frac{8}{3}ln5 - \frac{1}{3}ln2 + \frac{2}{3}\int_{1}^{2}(x^2+1-\frac{1}{x^2+1})dx[/TEX]
[TEX]A=\frac{8}{3}ln5 - \frac{1}{3}ln2 + \frac{2}{3}\int_{1}^{2}(x^2+1)dx - \frac{2}{3}\int_{1}^{2}(\frac{1}{1+x^2})dx[/TEX]
[TEX]A=\frac{8}{3}ln5 - \frac{1}{3}ln2 + \frac{2}{3}(\frac{8}{3}+2) -\frac{2}{3}(\frac{1}{3}+1) - \frac{2}{3}\int_{1}^{2}(\frac{1}{x^2+1})dx[/TEX]
[TEX]A=\frac{8}{3}ln5 - \frac{1}{3}ln2 + \frac{20}{9} -\frac{2}{3}A1[/TEX] (2)
[TEX]A1=\int_{1}^{2}\frac{dx}{x^2+1}[/TEX]
Đặt x= tant
[TEX]dx= (tant)^2 +1 dt[/TEX]
Đổi cận x = 1 => t = pi/4
x= 2 => t= số xấu ( làm tròn đại là 63 độ => 7pi/20)
[TEX]A1=\int_{\frac{pi}{4}}^{\frac{7pi}{20}}\frac{1+(tant)^2}{1+(tant)^2}dt[/TEX]
[TEX]A1=\int_{\frac{pi}{4}}^{\frac{7pi}{20}}dt[/TEX]
[TEX]A1=\frac{7pi}{20}-\frac{pi}{4}[/TEX]
[TEX]A1=\frac{pi}{10}[/TEX]
Thế A1 vào (2) => A = [TEX]\frac{8}{3}ln5 - \frac{1}{3}ln2 + \frac{20}{9} - \frac{pi}{15} [/TEX]
[TEX]B=\int_{1}^{2}xln(1+x^2)dx[/TEX]
Đặt
[TEX]u1 = ln(1+x^2) => du1= \frac{2x}{1+x^2}[/TEX]
[TEX]dv1=xdx => v1= \frac{x^2}{2}[/TEX]
[TEX]B= 2ln5 - \frac{1}{2}ln2 - \int_{1}^{2}\frac{x^3}{x^2+1}dx[/TEX]
[TEX]B= 2ln5 - \frac{1}{2}ln2 - \int_{1}^{2}(x- \frac{x}{x^2+1})dx[/TEX]
[TEX]B=2ln5 - \frac{1}{2}ln2 - \int_{1}^{2}xdx + \int_{1}^{2}(x^2+1)dx[/TEX]
[TEX]B=2ln5 - \frac{1}{2}ln2 - \frac{3}{2} + B1[/TEX] (3)
B1= [TEX] \int_{1}^{2}(x^2+1)dx [/TEX]
Đặt
[TEX]t=x^2 +1[/TEX]
[TEX]dt=2xdx[/TEX]
[TEX]\frac{dt}{2}=xdx[/TEX]
Đổi cận x= 1 => t = 2
x= 2 => t = 5
B1=[TEX]\int_{2}^{5}\frac{1}{2}\frac{dt}{t}[/TEX]
B1=[TEX]\frac{1}{2}ln5 - \frac{1}{2}ln2[/TEX]
Thế B1 vào (3) => B= [TEX]2ln5 - \frac{1}{2}ln2 - \frac{3}{2} + \frac{1}{2}ln5 - \frac{1}{2}ln2[/TEX]
Thế A , B vào (1)
I = [TEX]\frac{8}{3}ln5 - \frac{1}{3}ln2 + \frac{20}{9}- \frac{pi}{15}+ 2ln5 - \frac{1}{2}ln2 - \frac{3}{2} + \frac{1}{2}ln5 - \frac{1}{2}ln2[/TEX]
[TEX]I=\frac{31}{6}ln5 - \frac{4}{3}ln2 + \frac{13}{18} - \frac{pi}{15}[/TEX]
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn