# Integration by recognition, explanation needed

#### 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.

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

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

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'.

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

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.

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

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

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'.

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

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

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

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

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

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

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

• Tweety

#### Tweety

If evaluating the above line only, Tweety...

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

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

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

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

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

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

Thank you,

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

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

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

so

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

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

Thank you,

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

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

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

so

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

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

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

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

• Tweety