Best answer:
I=Sdx/[xsqr(x^4-1)]
Let u=sqr(x^4-1), then u^2=x^4-1
=>2udu=4(x^3)dx
=>udu=2(x^3)dx
=>dx=udu/[2(u^2+1)^(3/4)]
I=Sudu/[2(u^2+1)u]=>
I=(1/2)Sdu/(u^2+1)
Let u=tanA, then du=sec^2(A)dA
=>
I=(1/2)Ssec^2(A)dA/sec^2(A)
=>
I=(1/2)SdA
=>
I=(1/2)A+C
=>
I=(1/2)tan^-1(x^4-1)+C (Ans.)
Look into the...
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Best answer: I=Sdx/[xsqr(x^4-1)]
Let u=sqr(x^4-1), then u^2=x^4-1
=>2udu=4(x^3)dx
=>udu=2(x^3)dx
=>dx=udu/[2(u^2+1)^(3/4)]
I=Sudu/[2(u^2+1)u]=>
I=(1/2)Sdu/(u^2+1)
Let u=tanA, then du=sec^2(A)dA
=>
I=(1/2)Ssec^2(A)dA/sec^2(A)
=>
I=(1/2)SdA
=>
I=(1/2)A+C
=>
I=(1/2)tan^-1(x^4-1)+C (Ans.)
Look into the method you suggested :
I=S(x^3)dx/[(x^4)sqr(x^4-1)]
Let x=1/u, then dx=-du/u^2
=>
I=-Sdu/[(1/u)sqr((1/u^4)-1))]
=>
I=-Sudu/sqr(1-u^4)
=>
I= -(1/2)Sd(u^2)/sqr[1-(u^2)^2]
Let y=u^2, then dy=d(u^2)
=
I=-(1/2)Sdy/sqr(1-y^2)
Let y=sinA, then dy=cosAdA
=>
I=(-1/2)ScosAdA/cosA
=>
I=(-1/2)A+C
=>
I=(-1/2)sin^-1(y)+C
=>
I=(-1/2)sin^-1(u^2)+C
=>
I=(-1/2)sin^-1(1/x^2)+C
it is workable too, but clumsy.