Hello, a little stuck on a question:
Integrate (x+1)(x+3)^5 u = (x+3)
I have ended up with 1/7(x+3)^7 - 1/3(x+3)^6 + K
but the answer is 1/21 (x+3)^6(3x+2) ...
Can someone show me how to do it please
Thanks!
i must have made a mistake somewhere else because although thats very useful and what i was missing (thank you ) the answer i now get is:
21(x+3)^6(3x+2)
whereas the answer is:
$\displaystyle \frac{1}{21} (x+3)^6(3x+2)$ ... ??
$\displaystyle \frac{(x+3)^7}{7} - \frac{(x+3)^6}{3} $
$\displaystyle (x+3)^6 \{ \frac{(x+3)}{7} - \frac{1}{3} \} $
$\displaystyle (x+3)^6 \{ \frac{3(x+3)}{7 \cdot 3} - \frac{7}{3 \cdot 7 } \} $
$\displaystyle (x+3)^6 \{ \frac{3(x+3)-7}{21} \} $
$\displaystyle (x+3)^6 \{ \frac{3x+9-7}{21} \} $
$\displaystyle (x+3)^6 \{ \frac{(3x+2)}{21} \} $
That's it!
Feel free to ask if you still find it difficult ..
Hi darksupernova,
Using your substitution u=x+3, so x+1=u-2, du=dx
$\displaystyle \int{(u-2)u^5}du=\int{\left(u^6-2u^5\right)}du=\frac{u^7}{7}-\frac{u^6}{3}+C$
$\displaystyle =\frac{3u^7-7u^6}{21}+C$
$\displaystyle =\frac{u^6(3u-7)}{21}+C=\frac{(x+3)^6[3(x+3)-7]}{21}+C$
$\displaystyle =\frac{(x+3)^6(3x+9-7)}{21}+C=\frac{(x+3)^6(3x+2)}{21}+C$