∫1/1+√2x+1dx求不定积分
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∫1/1+√2x+1dx求不定积分
![∫1/1+√2x+1dx求不定积分](/uploads/image/z/17771952-48-2.jpg?t=%E2%88%AB1%2F1%2B%E2%88%9A2x%2B1dx%E6%B1%82%E4%B8%8D%E5%AE%9A%E7%A7%AF%E5%88%86)
∫1/(1+√(2x+1))dx
√(2x+1)=t (2x+1)=t^2 2dx=2tdt
∫1/(1+√(2x+1))dx
=∫tdt/(1+t)
=∫(t+1-1)dt/(1+t)
=∫(1-1/(1+t))dt
=t-ln(1+t)+C
=√(2x+1)-ln(1+√(2x+1))+C
√(2x+1)=t (2x+1)=t^2 2dx=2tdt
∫1/(1+√(2x+1))dx
=∫tdt/(1+t)
=∫(t+1-1)dt/(1+t)
=∫(1-1/(1+t))dt
=t-ln(1+t)+C
=√(2x+1)-ln(1+√(2x+1))+C