求定积分f(x)=1+∫[1,1/3](arctan√x)/(√x+√(x^3)dx求解,

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求定积分f(x)=1+∫[1,1/3](arctan√x)/(√x+√(x^3)dx求解,
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求定积分f(x)=1+∫[1,1/3](arctan√x)/(√x+√(x^3)dx求解,
求定积分f(x)=1+∫[1,1/3](arctan√x)/(√x+√(x^3)dx求解,

求定积分f(x)=1+∫[1,1/3](arctan√x)/(√x+√(x^3)dx求解,

如下

f(x)=1+∫[1,1/3](arctan√x)/(√x+√(x^3)dx
=1+∫[1,1/3](arctan√x)/(√x+√(x^3)d(√x)^2
=1+∫[1,1/3]2(arctan√x)/(1+√x^2)d(√x)
=1+(arctan√x)^2|(1/3,1)
=1+(π^2/16-π^2/36)
=