作业帮0 到无穷大 dx/((2+x^2)^2),
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0 到无穷大 dx/((2+x^2)^2),
0 到无穷大 dx/((2+x^2)^2),
0 到无穷大 dx/((2+x^2)^2),
∫ 1/((2+x^2)^2)dx
令x=√2tanu,则2+x^2=2(secu)^2,dx=√2(secu)^2du,x：0-->+无穷,u：0-->π/2
原式=∫ 1/(secu)^4*√2(secu)^2du
=√2∫ (cosu)^2du
=√2/2∫ (1+cos2u)du
=√2/2 (u+1/2sin2u) u：0-->π/2
=√2/2*π/2
=√2π/4
=(2^(1/2)*atan((2^(1/2)*x)/2))/8 + x/(4*(x^2 + 2))|(0,无穷大)
=(π*2^(1/2))/16
I=∫(0 to ∞)(1/2)[2+x^2-2x^2+x^2]dx/(2+x^2)^2
=(1/2)x/(2+x^2)|(0 to ∞)+∫(0 to ∞)(1/2)x^2dx/(2+x^2)^2
=(1/2)[0-0]+∫(0 to ∞)(1/2)[x^2+2-2]dx/(2+x^2)^2
=∫(0 to ∞)(1/2)dx/(2+x^2)-(1/2)∫(0 to ∞)...
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I=∫(0 to ∞)(1/2)[2+x^2-2x^2+x^2]dx/(2+x^2)^2
=(1/2)x/(2+x^2)|(0 to ∞)+∫(0 to ∞)(1/2)x^2dx/(2+x^2)^2
=(1/2)[0-0]+∫(0 to ∞)(1/2)[x^2+2-2]dx/(2+x^2)^2
=∫(0 to ∞)(1/2)dx/(2+x^2)-(1/2)∫(0 to ∞)2dx/(2+x^2)^2
=(1/2√2)arctan(x/√2)|(0 to ∞)-I
I=(1/4√2)arctan(x/√2)|(0 to ∞)
=(1/4√2)[π/2-0]
=π/(8√2)
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L = ∫(0->+∞) dx/(2+x²)²
设x = (√2)tany，dx = (√2)sec²y dy
当x -> 0时y -> 0；当x -> +∞时y -> π/2
(2+x²)² = (2+2tan²y)² = (2sec²y)² = 4sec⁴y
L...
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L = ∫(0->+∞) dx/(2+x²)²
设x = (√2)tany，dx = (√2)sec²y dy
当x -> 0时y -> 0；当x -> +∞时y -> π/2
(2+x²)² = (2+2tan²y)² = (2sec²y)² = 4sec⁴y
L = ∫(0->π/2) (√2)sec²y/(4sec⁴y) dy
= (2/√2 * 1/4)∫(0->π/2) dy/sec²y
= [1/(2√2)]∫(0->π/2) cos²y dy
= [1/(4√2)]∫(0->π/2) (1+cos2y) dy
= [1/(4√2)][y + (1/2)sin(2y)]：(0->π/2)
= [1/(4√2)][π/2 + 0]
= π/(8√2)
收起