求第二题过程答案
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(2)
∫sin(x/5)dx
=5∫sin(x/5)d(x/5)
=-5cos(x/5)+C
(4)
∫(1-2x)²dx
=∫(2x-1)²dx
=½∫(2x-1)²d(2x-1)
=½·⅓·(2x-1)³+C
=(1/6)(2x-1)³+C
(6)
∫[1/√(2+x)]dx
=∫[1/√(x+2)]d(x+2)
=2√(x+2)+C
(8)
∫[3/(x-2)²]dx
=3∫[1/(x-2)²]d(x-2)
=-3/(x-2) +C
∫sin(x/5)dx
=5∫sin(x/5)d(x/5)
=-5cos(x/5)+C
(4)
∫(1-2x)²dx
=∫(2x-1)²dx
=½∫(2x-1)²d(2x-1)
=½·⅓·(2x-1)³+C
=(1/6)(2x-1)³+C
(6)
∫[1/√(2+x)]dx
=∫[1/√(x+2)]d(x+2)
=2√(x+2)+C
(8)
∫[3/(x-2)²]dx
=3∫[1/(x-2)²]d(x-2)
=-3/(x-2) +C
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