1. ## Quick Integration Question

Let's say you had the integration problem (5x+7)^20. You would let u=5x+7 and du=5dx and so .2du=dx, and so you would end up with the integration of u^20 times .2 du. You could then bring out the constant term .2, and you would end up with .2 times the integration of u^20 du.

However, I remember that there are times in which you would multiply one side of the equation by a constant, then multiply the other side by its reciprocal after integrating the terms inside the integral. When is this exactly done? I don't think you do this in u-substitution, but I distinctly remember some instances in which you multiply by the reciprocal, not by the normal constant.

2. Originally Posted by jaijay32
Let's say you had the integration problem (5x+7)^20. You would let u=5x+7 and du=5dx and so .2du=dx, and so you would end up with the integration of u^20 times .2 du. You could then bring out the constant term .2, and you would end up with .2 times the integration of u^20 du.

However, I remember that there are times in which you would multiply one side of the equation by a constant, then multiply the other side by its reciprocal after integrating the terms inside the integral. When is this exactly done? I don't think you do this in u-substitution, but I distinctly remember some instances in which you multiply by the reciprocal, not by the normal constant.
$\int (5x+7)^{20} \, dx$

$u = 5x+7$

$du = 5 \, dx$

$\frac{1}{5} \int (5x+7)^{20} \cdot 5 \, dx$

substitute ...

$\frac{1}{5} \int u^{20} \, du$

$\frac{1}{5} \cdot \frac{1}{21} u^{21} + C$

$\frac{1}{105} (5x+7)^{21} + C$

now ... where is your confusion?