plz solve the calculus...
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Likun45:
https://brainly.in/question/1065112
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dear
you first let x^3+4=z^2
now its defferentiate w.r.t x
hence
3x^2dx=2z.dz
now
dx=2z.dz/3x^2
put this value in integration
i.e. integ 2z.dz/x (3x^2)z
=2/3integ dz/x^3
=2/3 integ dz/(z^2-4)
=2/3 {1/4 ln(z-2)/(z+2)}
=1/6ln {root (x^3+4)-2}/{root (x^3+4)+2)}
now I hope you integrate this
you first let x^3+4=z^2
now its defferentiate w.r.t x
hence
3x^2dx=2z.dz
now
dx=2z.dz/3x^2
put this value in integration
i.e. integ 2z.dz/x (3x^2)z
=2/3integ dz/x^3
=2/3 integ dz/(z^2-4)
=2/3 {1/4 ln(z-2)/(z+2)}
=1/6ln {root (x^3+4)-2}/{root (x^3+4)+2)}
now I hope you integrate this
please mark as brainliest
how??
let z = 2 sec y.... Then it will be of the form of integral cosec y dy ... you know the standard answer or derivation for that...
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