令t=√(6+5x),dx=d(t²-6)/5=2t/5dt,
=∫(√6到√11)2t/5(t+1)dt
=2/5∫1-1/(t+1)dt
=2/5(t-ln(t+1))
=2/5(√11-√6-ln(√11+1)+ln(√6+1))
三角换元脱根号,令x=secu,dx=tanusecudu,
=∫(0到π/4)tanu/secudsecu
=∫tan²udu
=tanu-u
=1-π/4
令x=tanu,
=∫(π/4到π/3)1/tan²usecudtanu
=∫secu/tan²udu
=∫cotucscudu
=-cscu
=2/√2-2/√3