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Integrate using U substitution only$$\int x(5x-1)^{19}$$I used $u=x^2$, and got $$ \frac{x(5x-1)^{20}}{100} + C$$which happend to be wrong. Please help, I'm stuck.

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  • $\begingroup$ The fact that you still have an integral sign in what you say is your answer suggests you should be more careful about some things. $\endgroup$ – Michael Hardy Jun 4 '14 at 2:32
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Try using the substitution $$u = 5x - 1.$$ instead

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Hint. If you had a (slightly) complicated linear term and just a single variable to a high power, for example, $$\int\Bigl(\frac{4x+5}{6}\Bigr)x^{78}\,dx\ ,$$ it would be easy. (How?) So, try substituting $u=5x-1$.

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No, use this: $u = 5x - 1$, and you will get

$$ \int\,x(5x - 1)^{19}\,dx = \int\, \left( \dfrac{u - 1}{5} \right)u^{19}\,du $$

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  • $\begingroup$ x = (u+1)/5 not (u-1)/5... $\endgroup$ – DeepSea Jun 4 '14 at 2:35
  • $\begingroup$ Yeah, you're right, sorry for that. $\endgroup$ – Alexei0709 Jun 4 '14 at 2:37
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Write: $x(5x-1)^{19} = \dfrac{1}{5}\cdot 5x(5x-1)^{19} = \dfrac{1}{5}\cdot ((5x-1) + 1)(5x-1)^{19} = \dfrac{1}{5}\cdot \left((5x-1)^{20} + (5x-1)^{19}\right) = \dfrac{(5x-1)^{20}}{5} + \dfrac{(5x-1)^{19}}{5}$, and you can make the substitution $u = 5x-1$, and get the anti-derivative.

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