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1.求根号下(a^2-x^2)^3的积分,所有内容都在根号内.2.(x+1)/(x^2+x+1)的积分
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1.求根号下(a^2-x^2)^3的积分,所有内容都在根号内.2.(x+1)/(x^2+x+1)的积分
▼优质解答
答案和解析
1,
令x=asint,则dx=acostdt
代入原式有
∫√(a^2-x^2)^3dt
=∫a^4(cost)^4dt
= a^4∫[(1+cos2t)/2]^2dt
= a^4/4∫[1+2cos2t+(1+cos4t)/2]dt
= a^4/4∫(3/2+2cos2t+1/2*cos4t)dt
= a^4/4(∫3/2dt+2∫cos2tdt+1/2∫cos4tdt)
= a^4/4(3t/2+sin2t+1/8*sin4t)+C
= a^4/4[3/2*arcsin(x/a)+2x√(a^2-x^2)/a^2+x(a^2-2x^2)√(a^2-x^2)/2a^4]+C
=3/8* arcsin(x/a) a^4+ 1/2*x√(a^2-x^2)a^2+1/8* x(a^2-2x^2)√(a^2-x^2)+C
2
∫(x+1)/(x^2+x+1)dx
=1/2∫(2x+2)/(x^2+x+1)dx
=1/2∫(2x+1)/(x^2+x+1)dx+1/2∫1/(x^2+x+1)dx
=1/2∫1/(x^2+x+1)d(x^2+x+1)+1/2∫1/[(x+1/2)^2+3/4)]dx
=1/2*ln(x^2+x+1)+ √3/3∫1/{[2/√3*(x+1/2)]^2]+1}d[2/√3*(x+1/2)]
=1/2*ln(x^2+x+1)+ √3/3arctan[2/√3*(x+1/2)]+C
令x=asint,则dx=acostdt
代入原式有
∫√(a^2-x^2)^3dt
=∫a^4(cost)^4dt
= a^4∫[(1+cos2t)/2]^2dt
= a^4/4∫[1+2cos2t+(1+cos4t)/2]dt
= a^4/4∫(3/2+2cos2t+1/2*cos4t)dt
= a^4/4(∫3/2dt+2∫cos2tdt+1/2∫cos4tdt)
= a^4/4(3t/2+sin2t+1/8*sin4t)+C
= a^4/4[3/2*arcsin(x/a)+2x√(a^2-x^2)/a^2+x(a^2-2x^2)√(a^2-x^2)/2a^4]+C
=3/8* arcsin(x/a) a^4+ 1/2*x√(a^2-x^2)a^2+1/8* x(a^2-2x^2)√(a^2-x^2)+C
2
∫(x+1)/(x^2+x+1)dx
=1/2∫(2x+2)/(x^2+x+1)dx
=1/2∫(2x+1)/(x^2+x+1)dx+1/2∫1/(x^2+x+1)dx
=1/2∫1/(x^2+x+1)d(x^2+x+1)+1/2∫1/[(x+1/2)^2+3/4)]dx
=1/2*ln(x^2+x+1)+ √3/3∫1/{[2/√3*(x+1/2)]^2]+1}d[2/√3*(x+1/2)]
=1/2*ln(x^2+x+1)+ √3/3arctan[2/√3*(x+1/2)]+C
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