Question

求不定积分x^2*(x^2+ 1)^0.5*dx

要详细过程

Answer 1

令x=tant dx=sec²tdt

∫ x^2*(x^2+ 1)^0.5*dx=∫tan²t sec³tdt=∫（sect）^5-sec³tdt

可再查查sect的n次方的递推公式，带入即可
答案是 （1/4）x*(x^2+1)^(3/2)-1/8*x*(x^2+1)^(1/2)-1/8*ln【x + (x^2+1)^(1/2)】+C
也可以如此做有理积分
∫tan²t sec³tdt=∫sin²t /(cost)^5dt =∫【sin²t /(1-sin²t)³】dsint
=∫ [u² /(1-u²)³] du

用待定系数法可得u² /(1-u²)³=(1/16)/(u-1)-(1/16)/(u-1)²-(1/8)/(u-1)³-(1/16)/(u+1)-(1/16)/(u+1)²-(1/8)/(u+1)³
∫ [u² /(1-u²)³] du =(1/16)ln|u-1|+(1/16)/(u-1)+(1/16)/(u-1)²-(1/16)ln(u+1)+(1/16)/(u+1)+(1/16)/(u+1)²+C

不过化简较麻烦

Answer 2

令x = tant，dx = sec^2t dt
∫ x^2√(x^2 + 1) dt
= ∫ tan^2t * sect * sec^2t dt
= ∫ (sec^2t - 1)sec^3t dt
= ∫ sec^5t dt - ∫ sec^3t dt
由公式∫ (sect)^n dt = [tant(sect)^(n - 2)]/(n - 1) + [(n - 2)/(n - 1)]∫ (sect)^(n - 2) dt
= (1/4)tant(sect)^3 + (3/4)∫ sec^3t dt - ∫ sec^3t dt
= (1/4)tant(sect)^3 - (1/4)∫ sec^3t dt
= (1/4)tant(sect)^3 - (1/4)[(1/2)secttant + (1/2)ln|sect + tant|] + C
= (1/4)tant(sect)^3 - (1/8)secttant - (1/8)ln|sect + tant| + C
= (x/4)(x^2 + 1)^(3/2) - (x/8)√(x^2 + 1) - (1/8)ln|x + √(x^2 + 1)| + C
= (x/8)(2x^2 + 1)√(x^2 + 1) - (1/8)ln|x + √(x^2 + 1)| + C