How to calculate the indefinite integral of 1x23(1+x23)? I substituted, t=1x13 dtdx=−13x43 dtdx=−t43 Rewriting the question, ∫dxx23+x43 −13∫dtt4(1t2+1t4) We have, −13∫dtt2+1 −13tan−1t+C −13tan−1(1x13)+C But the answer given is 3tan−1x13+C Where am I wrong?

Question

How to calculate the indefinite integral of 1x23(1+x23)?  I substituted,  t=1x13   dtdx=−13x43   dtdx=−t43   Rewriting the question,   ∫dxx23+x43   −13∫dtt4(1t2+1t4)   We have,   −13∫dtt2+1   −13tan−1t+C   −13tan−1(1x13)+C  But the answer given is   3tan−1x13+C   Where am I wrong?

Alfred Mueller · Accepted Answer

Two observations resolve this. The first is that you didn&#039;t invert the derivative properly, so your 1/3 coefficient should be 3 because dx=−3t−4dt. Secondly, arctan⁡1y=π2−arctan⁡y implies there&#039;s more than one way to write the answer, with your method getting something valid. I think you were expected to substitute t=x13 instead; your choice does work, but it&#039;s conceptually more complex.

utgyrnr0 · Accepted Answer

t=x3,dt=1x23dx  ∫1x23+x43dx=3∫1t2+1dt

How to calculate the indefinite integral of \frac{1}{x^{2/3}(1+x^{2/3})}? I substituted, t=\frac{1}{x^{1/3}} \frac{dt}{dx} =

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