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I must be missing something here, but I feel like I'm getting two different solutions when I try to solve the following integral using different u values in the u substitution method:

$$\int \frac{1}{300+2t} dt$$

When I pull 1/2 out of the integral beforehand and say u=150+t, du=dt, I get:

$$\frac{1}{2}\int \frac{1}{u} du = \frac{1}{2}ln(u) + c = \frac{1}{2}ln(150+t) + c$$

But when I leave the 1/2 in the integral and say u=300+2t, du=2dt, I get:

$$\int \frac{1}{u} \frac{du}{2} = \frac{1}{2}ln(u) + c = \frac{1}{2}ln(300+2t) + c$$

What am I doing wrong to get two different solutions to the integral?

Dan
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  • Related: http://math.stackexchange.com/q/495159/, http://math.stackexchange.com/questions/33187/trig-integral-int-cosx-sinx-cosx-dx – Jonas Meyer Sep 22 '13 at 05:01

1 Answers1

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Note that $$\ln(300 + 2t) = \ln(150 + t) + \ln{2}$$ The extra term gets rolled into the arbitrary constant $c$.