∫√(2x^2+9)dx
=(1/√2)∫√(2x^2+9)d√2x
√2x=3tanu secu=√[(√2x/3)^2+1]=√(2x^2+9)/3
=(1/√2)∫9secu^3du
=(1/√2)[(1/2)√2x√(2x^2+9)+(9/2)ln|√2x/3+√(2x^2+9)/3|]+C0
=(1/√2)[(1/2)√2x√(2x^2+9)+(9/)ln|√2x+√(2x^2+9)| ] +C
C=C0-(9/2)ln3
∫secu^3du=∫secudtanu=tanusecu-∫tanudsecu
=tanusecu-∫secu(tanu)^2du
=tanusecu-∫secudtanu+∫secudu
∫secu^3du=(1/2)tanusecu-(1/2)ln|secu+tanu|+C
