# Math Help - Integration by recognition, explanation needed

1. ## Integration by recognition, explanation needed

I am trying to understand the example in my book, however can't seem to makc sense of it. If someone could just explain it to me.

$\frac{d}{dx}(3x-2)^{5} = 15(3x-2)^{4}$

$\int 15(3x-2)^{4} dx = (3x-2)^{5} + C$

the next line I don't get, why do they divide both sides by 15? What's the need?

the book says 'an adjustment factor of 1/15 is needed'.

$\int (3x-2)^{4} dx = \frac{1}{15}(3x-2)^{5} + c$

2. Originally Posted by Tweety
I am trying to understand the example in my book, however can't seem to makc sense of it. If someone could just explain it to me.

$\frac{d}{dx}(3x-2)^{5} = 15(3x-2)^{4}$

$\int 15(3x-2)^{4} dx = (3x-2)^{5} + C$

the next line I don't get, why do they divide both sides by 15? What's the need?

the book says 'an adjustment factor of 1/15 is needed'.

$\int (3x-2)^{4} dx = \frac{1}{15}(3x-2)^{5} + c$
If evaluating the above line only, Tweety...

$\int{(3x-2)^4}dx=\frac{1}{15}(3x-2)^5+C$

because of $\frac{d}{dx}\left[(3x-2)^5+C\right]=15(3x-2)^4$

hence $\frac{1}{15}\ \frac{d}{dx}\left[(3x-2)^5+C\right]=(3x-2)^4$

therefore $\int{(3x-2)^4}dx=\frac{1}{15}\left[(3x-2)^5+C\right]$

You can try it using substitution of course... $u=3x-2,\ du=3dx,\ \frac{du}{3}=dx$

$\int{u^4}\frac{du}{3}=\frac{1}{3}\int{u^4}du=\frac {1}{3}\ \frac{1}{5}\ u^5+c=\frac{1}{15}(3x-2)^5+C$

3. Originally Posted by Archie Meade
If evaluating the above line only, Tweety...

$\int{(3x-2)^4}dx=\frac{1}{15}(3x-2)^5+C$

because of $\frac{d}{dx}\left[(3x-2)^5+C\right]=15(3x-2)^4$

hence $\frac{1}{15}\ \frac{d}{dx}\left[(3x-2)^5+C\right]=(3x-2)^4$

therefore $\int{(3x-2)^4}dx=\frac{1}{15}\left[(3x-2)^5+C\right]$

You can try it using substitution of course... $u=3x-2,\ du=3dx,\ \frac{du}{3}=dx$

$\int{u^4}\frac{du}{3}=\frac{1}{3}\int{u^4}du=\frac {1}{3}\ \frac{1}{5}\ u^5+c=\frac{1}{15}(3x-2)^5+C$

Thank you,

However I am still unclear about this, because we are trying to work out the $\int 15(3x-2)^{4} dx$

not $\int (3x-2)^{4} dx$

for example, $\frac{d}{dx} sin3x = cos3x \times 3$

so

$\int cos3x \times 3dx = sin3x + c$

so shouldn't it just be $\int 15(3x-2)^{4} dx = (3x-2)^{5} + C$ ?

4. Originally Posted by Tweety
Thank you,

However I am still unclear about this, because we are trying to work out the $\int 15(3x-2)^{4} dx$

not $\int (3x-2)^{4} dx$

for example, $\frac{d}{dx} sin3x = cos3x \times 3$

so

$\int cos3x \times 3dx = sin3x + c$

so shouldn't it just be $\int 15(3x-2)^{4} dx = (3x-2)^{5} + C$ ?
Hi Tweety,

I think the book example is calculating $\int{(3x-2)^4}dx$ using the result from $\int{15(3x-2)^4}dx$

since $\int{15(3x-2)^4}dx=15\int{(3x-2)^4}dx$