设(sinx2)'=f(x),∫f(x)dx=?不定积分∫1/(x+x²)dx=?
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![设(sinx2)'=f(x),∫f(x)dx=?不定积分∫1/(x+x²)dx=?](/uploads/image/z/3913899-51-9.jpg?t=%E8%AE%BE%28sinx2%29%27%3Df%28x%29%2C%E2%88%ABf%28x%29dx%3D%3F%E4%B8%8D%E5%AE%9A%E7%A7%AF%E5%88%86%E2%88%AB1%2F%28x%2Bx%26%23178%3B%29dx%3D%3F)
设(sinx2)'=f(x),∫f(x)dx=?不定积分∫1/(x+x²)dx=?
设(sinx2)'=f(x),∫f(x)dx=?不定积分∫1/(x+x²)dx=?
设(sinx2)'=f(x),∫f(x)dx=?不定积分∫1/(x+x²)dx=?
f(x)=cosx² * 2x
∫f(x)dx=∫cosx²dx²=sinx²+C
原式=∫{1/√[(x+1/2)²+3/4]}dx.
令x+1/2=(√3/2)tant,则:tant=(2x+1)/√3,dx=(√3/2)(sect)²dt,
且sint=√{(sint)²/[(sint)²+(cost)²]}=√{(tant)²/[1+(tant)²]}
=tant/√[1+(tant)²]=[(2x+1)/√3]/√[1+(2x+1)²/3]
=(2x+1)/√[3+(2x+1)²].
∴原式=∫{1/√[(3/4)(tant)²+3/4]}(√3/2)(sect)²dt
=∫(1/sect)(sect)²dt
=∫(1/cost)dt
=∫[1/(cost)²]d(sint)
=∫{1/[1-(sint)²]}d(sint)
=(1/2)∫{[(1+sint)+(1-sint)]/[(1+sint)(1-sint)]}d(sint)
=(1/2)∫[1/(1-sint)]d(sint)+(1/2)∫[1/(1+sint)]d(sint)
=(1/2)ln|1+sint|-(1/2)ln|1-sint|+C
=(1/2)ln|1+(2x+1)/√[3+(2x+1)²]|-(1/2)ln|1-(2x+1)/√[3+(2x+1)²]|+C
=(1/2)ln|√[3+(2x+1)²]+2x+1|-(1/2)ln|√[3+(2x+1)²-2x-1]+C
(sinx2)'=f(x),∫f(x)dx=sinx2 +C