Question

用第一换元法求不定积分 1，∫3x^2/x^6+1 dx 2，∫dx/1+9x^2 3, ∫dx/x^2+4x+13 帮个忙把过程写具体点吧

1，∫3x^2/x^6+1 dx
2，∫dx/1+9x^2
3, ∫dx/x^2+4x+13

Answer 1

附图

Answer 2

1，
∫3x^2/x^6+1 dx
=∫1/x^6+1 [3x^2dx]
=∫1/[(x^3)^2 +1] [dx^3]
= arctan(x^3) + C

2，
∫dx/(1+9x^2)
=1/3∫d(3x)/(1+(3x)^2)
=1/3arctan(3x)+C

3,
∫dx/x^2+4x+13
=∫dx/[(x+2)^2+9]
=1/9∫dx/{[(x+2)/3]^2+1}
=1/3∫[d(x+2)/3]/{[(x+2)/3]^2+1}
=1/3arctan[(x+2)/3] + C