作业帮计算积分,
来源：学生作业帮助网 编辑：作业帮 时间：2024/04/29 03:14:22
计算积分,
求定积分：[-1,1]∫[(2x²+xcosx)/(1+x²)]dx
原式=[-1,1]∫[2x²/(1+x²)]dx+[-1,1]∫[xcosx/(1+x²)]dx
由于第二个积分中的被积函数f(x)=xcosx/(1+x²)是一个奇函数：
f(-x)=-xcos(-x)/(x+x²)=-xcosx/(1+x²)=-f(x),其在对称区间上的积分=0,故
原式=[-1,1]∫[2x²/(1+x²)]dx=[-1,1]2∫[1-1/(1+x²)]dx=2[x-arctanx]︱[-1,1]=2[(1-π/4)-(-1+π/4)]
=2(2-π/2)=4-π
∫(2x^2+xcosx)dx/(1+x^2)
=∫2x^2dx/(1+x^2)+∫xcosxdx/(1+x^2)
=2x-2∫dx/(1+x^2)+∫xcosxdx/(1+x^2)
=2x-2arctanx+∫xcosxdx/(1+x^2)
∫[-1,1]xcosxdx/(1+x^2)=∫[-1,0]xcosxdx/(1+x^2) +∫[0,1]xcosxdx/(...
全部展开
∫(2x^2+xcosx)dx/(1+x^2)
=∫2x^2dx/(1+x^2)+∫xcosxdx/(1+x^2)
=2x-2∫dx/(1+x^2)+∫xcosxdx/(1+x^2)
=2x-2arctanx+∫xcosxdx/(1+x^2)
∫[-1,1]xcosxdx/(1+x^2)=∫[-1,0]xcosxdx/(1+x^2) +∫[0,1]xcosxdx/(1+x^2) u=-x
=∫[-1,0]xcosxdx/(1+x^2) + ∫[0,-1] -ucos(-u)d(-u)/(1+(-u)^2)
=∫[-1,0]xcosxdx/(1+x^2) +∫[0,-1]ucosudu/(1+u^2)
=∫[-1,-1]xcosxdx/(1+x^2)=0
∫[-1,1]xcosxdx/(1+x^2)
=2x-2arctanx |[-1,1]
=4-2(π/4-(-π/4)
=4-π
收起