Author Topic: Integration Cases (Read 3280 times) Tweet Share

kenhung123 · « **Reply #30 on:** July 14, 2010, 09:18:07 pm »

Why is it that $\int \frac{1}{5-2x}dx=\frac{-1}{2}log_e|2x-5|+C$ not $\int \frac{1}{5-2x}dx=\frac{-1}{2}log_e|5-2x|+C$ but $\int \frac{1}{(5-2x)^{2}}dx=\frac{1}{2}(5-2x)^{-1}+C$

tcg93 · « **Reply #31 on:** July 14, 2010, 09:19:43 pm »

Quote from: kenhung123 on July 14, 2010, 09:18:07 pm

Why is it that $\int \frac{1}{5-2x}dx=\frac{-1}{2}log_e|2x-5|+C$ not $\int \frac{1}{5-2x}dx=\frac{-1}{2}log_e|5-2x|+C$ but $\int \frac{1}{(5-2x)^{2}}dx=\frac{1}{2}(5-2x)^{-1}+C$

The mod makes them the same thing - try subbing any value of x in - you will get the same result. 2x - 5 just looks nicer.

kenhung123 · « **Reply #32 on:** July 15, 2010, 10:17:48 pm »

thanks

why is it that $\int \frac{1}{2x}dx\neq \frac{1}{x}log_e(2x)+C$ ?

TrueTears · « **Reply #33 on:** July 15, 2010, 10:18:38 pm »

Quote from: kenhung123 on July 15, 2010, 10:17:48 pm

thanks

why is it that $\int \frac{1}{2x}dx\neq \frac{1}{x}log_e(2x)+C$ ?

differentiate the RHS and you will see why

kenhung123 · « **Reply #34 on:** July 15, 2010, 10:35:20 pm »

Yea but why doesn't that formula apply? (ax+b)^-1

TrueTears · « **Reply #35 on:** July 15, 2010, 10:36:35 pm »

i dono that formula, i just do it manually.

Martoman · « **Reply #36 on:** July 15, 2010, 11:00:31 pm »

Quote from: kenhung123 on July 15, 2010, 10:17:48 pm

thanks

why is it that $\int \frac{1}{2x}dx\neq \frac{1}{x}log_e(2x)+C$ ?

Ok this is really.basic. but take out 1/2 then you have 1/x -> .5ln|x|

tcg93 · « **Reply #37 on:** July 16, 2010, 12:56:34 pm »

Quote from: kenhung123 on July 15, 2010, 10:35:20 pm

Yea but why doesn't that formula apply? (ax+b)^-1

yeah don't use the formula because you may "forget" it in the exam, just manipulate the integral moving coefficients outside the integral so you get a function's derivative on top of the function

Martoman · « **Reply #38 on:** July 16, 2010, 02:19:25 pm »

Quote from: kenhung123 on July 15, 2010, 10:35:20 pm

Yea but why doesn't that formula apply? (ax+b)^-1

it doesn't apply because $(ax+b)^{-1+1} = (ax+b)^0 = 1$ , a trivial case where the power rule fails.

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