Question

不定积分

∫ x^98/[(1-x^2)^(101/2)]dx
∫ ln(1+√[(1+x)/x]dx

Answer 1

第一题：

令x = sinθ，dx = cosθ dθ

∫ x⁹⁸/(1 - x²)¹º½ dx

= ∫ sin⁹⁸θcosθ/cos¹º¹θ dθ

= ∫ tan⁹⁸θsec²θ dθ

= ∫ tan⁹⁸θ d(tanθ)

= tan⁹⁹θ/99 + C

= (1/99)[x/√(1 - x²)]⁹⁹ + C

= x⁹⁹/[99(1 - x²)^(99/2)] + C

第二题：

Answer 2

∫x^98 dx/(1-x^2)^(101/2)
=∫x^98dx/[(x^100)√(1-x^2) *√(1/x^2-1)]
=∫dx/[x^2√(1-x^2)*√(1/x^2-1)]
=∫dx/[x^3 *√(1/x^2-1)]
=(-1/2)∫d(1/x^2 -1)/√(1/x^2 -1)
=-√(1/x^2 -1) +C

∫ln(1+√[(1+x)/x] )dx
=∫ln(√x+√(1+x) -ln√x dx
=∫ln(√x+√(1+x) dx -∫ln√xdx ln(√x+√(1+x) '=[1/2√x+1/2√(1+x)]/[√x+√(1+x)]=(1/2)(1/√x(1+x)]
=(1/2)∫xdx/√x(1+x)-∫xdx/(2√x)(√x)
=(1/2)∫√xdx/√(1+x) -(1/2)∫dx
=(1/2)√[x(1+x)]-(1/2)ln|√x+√(1+x)| -(1/2)x+C

∫√xdx/√(1+x) x=tanu^2 dx=2tanusecu^2 du
=∫2tanu^2secudu
=∫2tanudsecu
=2tanusecu-2∫secu^3du
=2tanusecu-2∫tanu^2secudu-2∫secudu
4∫tanu^2secudu=2tanusecu-2∫secudu=2tanusecu-2ln|secu+tanu|
2∫tanu^2secudu=tanusecu-ln|secu+tanu|=√[x(1+x)] -ln|√(1+x)+√x|