高数不定积分 f'(tanx)=sec^2x,且f(0)=1,求f(x)

答案是x+x^3/3+C，求详细过程，怎么算的

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ƒ'(tanx) = sec²x = 1 + tan²洞没x
ƒ'(x) = 1 + x²
ƒ(x) = ∫ (1 + x²) dx = x + x³/3 + C
ƒ(0) = 1 ==> 0 + 0/3 + C = 1 ==>升携 C = 1
∴ƒ(x) = x + x³/吵颤伏3 + 1

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用悔侍sec^2x在tanx下积分 =用碧卜吵csc^2x+1在tanx下积分 =(tan^3x)/弊羡3+tanx+C
f(tanx)=(tan^3x)/3+tanx+C f(0)=1 C=1
令x=tanx
f(x)=x+x^3/3+1

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