Question

xarctanxdx的定积分0到1是什么？

Answer 1

∫xarctanxdx=1/2 ∫arctanxdx^2

=1/2[x^2arctanx|(0,1)-∫(0,1)x^2/(1+x^2)dx]

=1/2[π/4-∫(0,1)1-1/(1+x^2)dx]

=1/2[π/4-∫(0,1)dx+∫(0,1)1/(1+x^2)dx]

=1/2[π/4-x|(0,1)+arctanx|(0,1)]

=π/4-1/2

积分基本公式

1、∫0dx=c

2、∫x^udx=(x^u+1)/(u+1)+c

3、∫1/xdx=ln|x|+c

4、∫a^xdx=(a^x)/lna+c

5、∫e^xdx=e^x+c

6、∫sinxdx=-cosx+c

7、∫cosxdx=sinx+c

8、∫1/(cosx)^2dx=tanx+c

9、∫1/(sinx)^2dx=-cotx+c