Question

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Answer 1

Answer:

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Answer 2

I=1/√x(1+x^1/3)…(1)

Put 1+x^1/3 =t

Differentiate w.r.t.x

0+1/3 x^-2/3 =dt/dx

Or 1/3x^2/3 =dt/dx

Or dx/√x =3xdt

From (1)

I=3xdt/t

=3(t-1)^3dt/t

=3[(t^3 -3t^2 +3t -1)dt/t]

=3[(t^2 -3t+3–1/t)dt]

integrate

I=3[t^3 /3 -3t^2 /2 +3t-ln|t|]+c

Where t=(1+x^1/3)