# Math Help - Integration by substitution

1. ## Integration by substitution

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!

2. You integrate the thing and then you will get

You will get

(x+3)^6 [ 1/7(x+3) - 1/3] + K

Any more help ?

3. 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)

$\frac{1}{21} (x+3)^6(3x+2)$ ... ??

4. Originally Posted by darksupernova
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)

1/21 (x+3)^6(3x+2) ... ??
$\frac{(x+3)^7}{7} - \frac{(x+3)^6}{3}$

$(x+3)^6 \{ \frac{(x+3)}{7} - \frac{1}{3} \}$

$(x+3)^6 \{ \frac{3(x+3)}{7 \cdot 3} - \frac{7}{3 \cdot 7 } \}$

$(x+3)^6 \{ \frac{3(x+3)-7}{21} \}$

$(x+3)^6 \{ \frac{3x+9-7}{21} \}$

$(x+3)^6 \{ \frac{(3x+2)}{21} \}$

That's it!

Feel free to ask if you still find it difficult ..

5. ahh... didnt think about doing that!! thanks!

6. Originally Posted by darksupernova
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!
Hi darksupernova,

Using your substitution u=x+3, so x+1=u-2, du=dx

$\int{(u-2)u^5}du=\int{\left(u^6-2u^5\right)}du=\frac{u^7}{7}-\frac{u^6}{3}+C$

$=\frac{3u^7-7u^6}{21}+C$

$=\frac{u^6(3u-7)}{21}+C=\frac{(x+3)^6[3(x+3)-7]}{21}+C$

$=\frac{(x+3)^6(3x+9-7)}{21}+C=\frac{(x+3)^6(3x+2)}{21}+C$