第1和3题求详细过程!谢谢
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解:
(1)
令√x=t,则x=t²
x:0→4,则t:0→2
∫[0:4] dx/(1+√x)
=∫[0:2] [1/(1+t)]d(t²)
=∫[0:2] [2t/(1+t)]dt
=∫[0:2] [(2t+2-2)/(1+t)]dt
=∫[0:2] [2 -2/(1+t)]dt
=∫[0:2]2dt -2∫[0:2][1/(1+t)]d(1+t)
=2t|[0:2] -2ln|1+t||[0:2]
=2(2-0)-2(ln3-ln1)
=4-2ln3
(3)
∫[0:1]x(x²+1)⁵dx
=½∫[0:1](x²+1)⁵d(x²+1)
=½·(1/6)(x²+1)⁶|[0:1]
=(1/12)[(1²+1)⁶-(0²+1)⁶]
=(1/12)(64-1)
=21/4
(1)
令√x=t,则x=t²
x:0→4,则t:0→2
∫[0:4] dx/(1+√x)
=∫[0:2] [1/(1+t)]d(t²)
=∫[0:2] [2t/(1+t)]dt
=∫[0:2] [(2t+2-2)/(1+t)]dt
=∫[0:2] [2 -2/(1+t)]dt
=∫[0:2]2dt -2∫[0:2][1/(1+t)]d(1+t)
=2t|[0:2] -2ln|1+t||[0:2]
=2(2-0)-2(ln3-ln1)
=4-2ln3
(3)
∫[0:1]x(x²+1)⁵dx
=½∫[0:1](x²+1)⁵d(x²+1)
=½·(1/6)(x²+1)⁶|[0:1]
=(1/12)[(1²+1)⁶-(0²+1)⁶]
=(1/12)(64-1)
=21/4
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