integrate $\displaystyle (x-1) / (2x+1)^3 dx $
i found that it will be $\displaystyle -1/2 (2x+1)^-^1 + 3/4 (2x+1)^-^2 +c $
but i have guide answer : $\displaystyle -1/4 (2x+1)^-^1 + 3/8 (2x+1)^-^2 +c $
integrate $\displaystyle (x-1) / (2x+1)^3 dx $
i found that it will be $\displaystyle -1/2 (2x+1)^-^1 + 3/4 (2x+1)^-^2 +c $
but i have guide answer : $\displaystyle -1/4 (2x+1)^-^1 + 3/8 (2x+1)^-^2 +c $
Uhm...
$\displaystyle \int \frac{x-1}{2x+1}\, dx=\int \frac{1}{2}\cdot \frac{2(x-1)}{2x+1}\, dx=$
$\displaystyle =\frac{1}{2}\int \frac{2x+1-3}{2x+1}\, dx=\frac{1}{2}\int \left (1-\frac{3}{2x+1} \right )\, dx=$
$\displaystyle =\frac{1}{2}\left (\int 1\, dx -\int\frac{3}{2} \cdot \frac{2}{2x+1} \right )\, dx \right )=$
$\displaystyle =\frac{1}{2}\left ( x- \frac{3}{2} \ln (2x+1)\right )+C$
*her ^^'but i can't understand the last two steps in his solution how 2/(2x+1) become (2x+1) ???
Do you know the integration/derivative rules? If not, you should first learn them.
$\displaystyle \int \frac{x-1}{(2x+1)^3}\, dx=\int \frac{1}{2}\cdot \frac{2(x-1)}{(2x+1)^3}\, dx=$
$\displaystyle =\frac{1}{2}\int \frac{2x+1-3}{(2x+1)^3}\, dx=\frac{1}{2}\int \left (\frac{1}{(2x+1)^2}-\frac{3}{(2x+1)^3} \right )\, dx=$
$\displaystyle =\frac{1}{2}\left (\int (2x+1)^{-2}\, dx -3\int (2x+1)^{-3} \right )\, dx \right )=$
$\displaystyle =\frac{1}{2}\left (\int \frac{1}{2} \left (- \frac{1}{2x+1} \right )' \, dx -3\int \frac{1}{4}\left ( \frac{1}{(2x+1)^2} \right )' \right )\, dx \right )=$
$\displaystyle =-\frac{1}{4}\cdot \frac{1}{2x+1}-\frac{3}{8}\cdot \frac{1}{(2x+1)^2}+C$
Their answer is the correct one. If that's all you wanted to know, you could have used WolframAlpha.