# Thread: substitution method question

1. ## substitution method question

Do can you tell when you can use the substitution method? I just tried it on $\int_{-1}^0(2x+3)^2dx$, but I don't get the right answer unless I square the $(2x+3)$ and then proceed normally.

What about substitution on $\int_3^2\frac{x^2-1}{x-1}dx$?

2. Originally Posted by cinder
Do can you tell when you can use the substitution method? I just tried it on $\int_{-1}^0(2x+3)^2dx$, but I don't get the right answer unless I square the $(2x+3)$ and then proceed normally.
What about substitution on $\int_3^2\frac{x^2-1}{x-1}dx$?
Hello,

to 1.:
use u(x) = 2x+3. Then du/dx = 2. That means du = 2*dx.

So you could use the substitution method if there are the factor 2:

$\int_{-1}^0 \frac{1}{2} \cdot 2 \cdot (2x+3)^2dx=\int_{-1}^0 \frac{1}{2} u^2 du$

$\int_{-1}^0 \frac{1}{2} u^2 du= \frac{1}{6}u^3+c$

Now you can re-substitute and you'll get:

$\int_{-1}^0 \frac{1}{2} \cdot 2 \cdot (2x+3)^2dx=\frac{1}{6} \cdot (2x+3)^3+c$

to 2.:
Factorize (x^2-1) = (x+1)(x-1). Then you can cancel (x-1) and you'll get:

$\int_3^2\frac{x^2-1}{x-1}dx= \int_3^2 (x+1)dx$.

From here on you know what to do!

Greetings

EB

3. Originally Posted by earboth
Hello,

to 1.:
use u(x) = 2x+3. Then du/dx = 2. That means du = 2*dx.

So you could use the substitution method if there are the factor 2:

$\int_{-1}^0 \frac{1}{2} \cdot 2 \cdot (2x+3)^2dx=\int_{-1}^0 \frac{1}{2} u^2 du$

$\int_{-1}^0 \frac{1}{2} u^2 du= \frac{1}{6}u^3+c$

Now you can re-substitute and you'll get:

$\int_{-1}^0 \frac{1}{2} \cdot 2 \cdot (2x+3)^2dx=\frac{1}{6} \cdot (2x+3)^3+c$

to 2.:
Factorize (x^2-1) = (x+1)(x-1). Then you can cancel (x-1) and you'll get:

$\int_3^2\frac{x^2-1}{x-1}dx= \int_3^2 (x+1)dx$.

From here on you know what to do!

Greetings

EB
Okay, so on the first one, could you do:

$\int_{-1}^0(2x+3)^2dx$

$u = (2x + 3)$
$du = 2dx$ which becomes $\frac{1}{2}du=dx$

then

$\int_1^3u^2\frac{1}{2}du = \frac{1}{2}\int_1^3u^2du$

and solve normally, just using $du$?

4. Originally Posted by cinder
and solve normally, just using $du$?
Yes, that what $du$ means, a function in terms of $u$ whose derivative gives you whatever the function is. Or another way of saying the antiderivative in terms of $u$. That is the entire point of substitution because it simplifys the integrand into a simpler one.

5. Originally Posted by ThePerfectHacker
Yes, that what $du$ means, a function in terms of $u$ whose derivative gives you whatever the function is. Or another way of saying the antiderivative in terms of $u$. That is the entire point of substitution because it simplifys the integrand into a simpler one.
Okay, did I do the substitution right?

6. Originally Posted by cinder
Okay, did I do the substitution right?
Yes

7. Originally Posted by ThePerfectHacker
Yes
Thanks.