∫(√(2x+1))/(x^2)dx
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∫(√(2x+1))/(x^2)dx
![∫(√(2x+1))/(x^2)dx](/uploads/image/z/19264185-9-5.jpg?t=%E2%88%AB%28%E2%88%9A%282x%2B1%29%29%2F%28x%5E2%29dx)
设√(2x+1)=t,则x=(t²-1)/2,dx=tdt
∴原式=4∫t²dt/(t²-1)²
=∫[1/(t-1)-1/(t+1)+1/(t+1)²+1/(t-1)²]dt
=ln│t-1│-ln│t+1│-1/(t+1)-1/(t-1)+C (C是积分常数)
=ln│(t-1)/(t+1)│-2t/(t²-1)+C
=ln│(√(2x+1)-1)/(√(2x+1)+1)│-2√(2x+1)/(2x)+C
=ln│(√(2x+1)-1)/(√(2x+1)+1)│-√(2x+1)/x+C.
∴原式=4∫t²dt/(t²-1)²
=∫[1/(t-1)-1/(t+1)+1/(t+1)²+1/(t-1)²]dt
=ln│t-1│-ln│t+1│-1/(t+1)-1/(t-1)+C (C是积分常数)
=ln│(t-1)/(t+1)│-2t/(t²-1)+C
=ln│(√(2x+1)-1)/(√(2x+1)+1)│-2√(2x+1)/(2x)+C
=ln│(√(2x+1)-1)/(√(2x+1)+1)│-√(2x+1)/x+C.