problem in integration

**mido22** · Feb 20th 2013, 03:03 AM

integrate $(x-1) / (2x+1)^3 dx$

i found that it will be $-1/2 (2x+1)^-^1 + 3/4 (2x+1)^-^2 +c$

but i have guide answer : $-1/4 (2x+1)^-^1 + 3/8 (2x+1)^-^2 +c$

**MathJack** · Feb 20th 2013, 03:40 AM

split the fraction so ( x/(2x+1) -1/(2x + 1))dx
now let u= 2x + 1
x =(u -1)/2
integrate separately using the above substitution on the first piece
the second is just -1/u integrated

**veileen** · Feb 20th 2013, 03:44 AM

Uhm...

$\int \frac{x-1}{2x+1}\, dx=\int \frac{1}{2}\cdot \frac{2(x-1)}{2x+1}\, dx=$

$=\frac{1}{2}\int \frac{2x+1-3}{2x+1}\, dx=\frac{1}{2}\int \left (1-\frac{3}{2x+1} \right )\, dx=$

$=\frac{1}{2}\left (\int 1\, dx -\int\frac{3}{2} \cdot \frac{2}{2x+1} \right )\, dx \right )=$

$=\frac{1}{2}\left ( x- \frac{3}{2} \ln (2x+1)\right )+C$

**mido22** · Feb 20th 2013, 04:26 AM

sorry all but there is power 3 on (2x+1) in the problem

**MathJack** · Feb 20th 2013, 04:31 AM

That doesnt make much of a difference, use veileens method with the power of 3

**mido22** · Feb 20th 2013, 04:48 AM

Originally Posted by MathJack

That doesnt make much of a difference, use veileens method with the power of 3

but i can't understand the last two steps in his solution how 2/(2x+1) become (2x+1) ???
sorry but this is my first day to learn about integration ....

**mido22** · Feb 20th 2013, 04:49 AM

either my solution correct or the guide answer is correct??

**veileen** · Feb 20th 2013, 09:01 AM

but i can't understand the last two steps in his solution how 2/(2x+1) become (2x+1) ???

*her ^^'
Do you know the integration/derivative rules? If not, you should first learn them.

$\int \frac{x-1}{(2x+1)^3}\, dx=\int \frac{1}{2}\cdot \frac{2(x-1)}{(2x+1)^3}\, dx=$

$=\frac{1}{2}\int \frac{2x+1-3}{(2x+1)^3}\, dx=\frac{1}{2}\int \left (\frac{1}{(2x+1)^2}-\frac{3}{(2x+1)^3} \right )\, dx=$

$=\frac{1}{2}\left (\int (2x+1)^{-2}\, dx -3\int (2x+1)^{-3} \right )\, dx \right )=$

$=\frac{1}{2}\left (\int \frac{1}{2} \left (- \frac{1}{2x+1} \right )' \, dx -3\int \frac{1}{4}\left ( \frac{1}{(2x+1)^2} \right )' \right )\, dx \right )=$

$=-\frac{1}{4}\cdot \frac{1}{2x+1}-\frac{3}{8}\cdot \frac{1}{(2x+1)^2}+C$

Their answer is the correct one. If that's all you wanted to know, you could have used WolframAlpha.

**MINOANMAN** · Feb 20th 2013, 09:09 AM

Mido22

expand the function to get 1/2(2x+1)^2 -1/2(2x+1)^3 and then integrate to get (1-4x)/8(2x+1)^2 +C
it is easy.
MINOAS

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