evaluate:f5/(6x+13+x^2)dxevaluate:f(2x+1)/(x+1)evaluate:fx((x^2)-1)^4dxevaluate:f(x^2)^（1/5）evaluate:f3(cscx)^2evaluate:f((x^4)-(x^3))/(x^2)evaluate:f（(tanx)(secx)^3）/(1+(tanx)^2)dxevaluate是求解的意思急用

Question

evaluate:f 5/(6x+13+x^2) dx
evaluate:f (2x+1)/(x+1)
evaluate:f x((x^2)-1)^4 dx
evaluate:f (x^2)^（1/5）
evaluate:f 3(cscx)^2
evaluate:f ((x^4)-(x^3))/(x^2)
evaluate:f（(tanx)(secx)^3）/(1+(tanx)^2) dx
evaluate 是求解的意思
急用

Answer 1

∫5dx/[(x+3)^2+4]=10∫d(x/2+3/2)/[(x/2+3/2)^2+1]=10arctan(x/2+3/2)+C
∫(2x+1)dx/(x+1)=2x-ln(x+1)+C
∫xdx/(x^2-1)^4=(1/2)*(-1/3)[1/(x^2-1)^3]+C
∫(x^2)^(1/5)dx =∫x^(2/5)dx=(5/7)x^(7/5)+C
∫3(cscx)^2dx=3∫dx/(sinx)^2=3∫[(sinx)^2+(cosx)^2]dx/sinx^2
=-3∫(sinxdcosx-cosxdsinx)/(sinx)^2=-3(cosx/sinx)+C
∫(x^4-x^3)dx/x^2=∫x^2dx-∫xdx=x^3/3-x^2/2+C
∫tanx(secx)^3 dx/[1+(tanx)^2]=∫tanx(secx)^3dx/(secx)^2=∫tanxsecxdx
=∫sinxdx/(cosx)^2=-∫dcosx/(cosx)^2=1/cosx+C=secx+C