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So I'm given this example in my notes:

$$\int x(x+3)^5dx$$

And when I integrate by parts and by substitution I end up getting two different answers, but I don't know why.

Integrating by parts:
If I let $u = x$ and $dv = (x + 3)^5$, then:

$$ \begin{align*} \int x(x+3)^5dx &= {x \over 6 }(x+3)^6 - \int {1 \over 6 }(x+3)^6dx\\ &= {x \over 6 }(x+3)^6 - {1 \over 6 }({1 \over 7 }(x+3)^7) + C\\ &= {x \over 6 }(x+3)^6 - {1 \over 42 }(x+3)^7 + C \end{align*} $$

Integration by u-substitution:
If I let $u = x + 3$, then ${du\over dx } = 1$ and $du = dx$: $$ \begin{align*} \int x(x+3)^5dx &= \int (u-3)u^5 \ du\\ &= \int u^6 - 3u^5\ du\\ &= {(x+3)^7 \over 7 } - {(x+3)^6 \over 2 } + C \end{align*} $$

Where am I going wrong?

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    $\begingroup$ In the future if you ever want to really see if the answers are equivalent, you can take the derivative of both and see if you get your original function and/or you could graph both answers (the worse that could happen here is that they're shifted up or down from each other by a constant) $\endgroup$ – user12345 Jun 5 '17 at 19:56
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In fact, the two solutions are equivalent (if you put the forgotten $C$ on the first integral): \begin{align*} \frac{x}{6}(x+3)^6 - \frac{1}{42}(x+3)^7 + C & = \frac{(x+3)-3}{6}(x+3)^6 - \frac{1}{42}(x+3)^7 + C \\ & = \frac{1}{6} (x+3)^7 - \frac{1}{2} (x+3)^6 - \frac{1}{42} (x+3)^7 + C \\ & = \frac{1}{7} (x+3)^7 - \frac{1}{2} (x+3)^6 + C. \end{align*}

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