Evaluate the following integral ∫1/x(x^5+3) dx
Put 1/z=x^5 so x=z^(-1/5) and dx=-1/5 z^(-6/5) dz∫1/(z^(-1/5) (1/z+3) )-1/5 z^(-6/5) dz=-1/5 ∫z^(-6/5) dzI=(-1)/15 ln|1+3z|=(-1)/15 ln|1+3x^(-5) |+c
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Put 1/z=x^5
need an explanation for this answer? contact us directly to get an explanation for this answerso x=z^(-1/5) and dx=-1/5 z^(-6/5) dz
∫1/(z^(-1/5) (1/z+3) )-1/5 z^(-6/5) dz=-1/5 ∫z^(-6/5) dz
I=(-1)/15 ln|1+3z|=(-1)/15 ln|1+3x^(-5) |+c