MHB Antidifferentiation by Substitution

[bunnypatotie]

1.\[ \int x^2 e^{x^3} dx \]
2. \[ \int sin(2x-3)dx \]
3. \[ \int (\cfrac {3dx}{(x+2)\sqrt {x^2+4x+3}} ) \]
4. \[ \int (\cfrac {x^3}{(x^2 +4)^\cfrac {3}{2}} )dx \]

 

[topsquark]

Where are you having a problem? Show us what you've been able to do with these and we can help you much better.

-Dan

 

[HallsofIvy]

Once you get more experience, I think you will be able to see the substitution for problems easily!

For 1, let $u= x^3$. Then what is du? Do you see that "$x^2$" in the integral? What happens to that?

For 2, let $u= 2x- 3$. Then what is du?

For 3, let $u= x^2+ 4x+ 3$. Then what is du?
Be careful about the fact that "x+ 2" is in the denominator!

For 4, let $u= x^2+ 4$. What is du? Notice that $x^2dx= x^2(xdx)$ and $x^2= u- 4$.

 

[bunnypatotie]

Once you get more experience, I think you will be able to see the substitution for problems easily!

For 1, let $u= x^3$. Then what is du? Do you see that "$x^2$" in the integral? What happens to that?

For 2, let $u= 2x- 3$. Then what is du?

For 3, let $u= x^2+ 4x+ 3$. Then what is du?
Be careful about the fact that "x+ 2" is in the denominator!

For 4, let $u= x^2+ 4$. What is du? Notice that $x^2dx= x^2(xdx)$ and $x^2= u- 4$.

I'm already done with numbers 1, 3 and 4. I'm stuck with no.2 since I have no idea how to solve it

 

[bunnypatotie]

Where are you having a problem? Show us what you've been able to do with these and we can help you much better.

-Dan

I'm done with items 1, 3 and 4. I'm stuck with item no. 2 since I don't have any idea how to solve this problem.

 

[bunnypatotie]

I'm already done with numbers 1, 3 and 4. I'm stuck with no.2 since I have no idea how to solve it

I meant no. 3 sorry. I'm done with nos. 1, 2 and 4 already. Haven't started with no. 3 yet because I don't have any idea how to really solve this problem

 

[skeeter]

$\displaystyle 3\int \dfrac{dx}{(x+2)\sqrt{x^2+4x+4-1}} = 3\int \dfrac{dx}{(x+2)\sqrt{(x+2)^2-1}}$

let $u = x+2 \implies du = dx$ ...

$\displaystyle 3\int \dfrac{1}{u\sqrt{u^2-1}} \, du$

substitution again ...

$v = \sqrt{u^2-1}$

continue ...

 

[bunnypatotie]

$\displaystyle 3\int \dfrac{dx}{(x+2)\sqrt{x^2+4x+4-1}} = 3\int \dfrac{dx}{(x+2)\sqrt{(x+2)^2-1}}$

let $u = x+2 \implies du = dx$ ...

$\displaystyle 3\int \dfrac{1}{u\sqrt{u^2-1}} \, du$

substitution again ...

$v = \sqrt{u^2-1}$

continue ...

Got it, tysm.

 

[HallsofIvy]

What skeeter did was "complete the square" $x^2+4x+ 3= x^3+ 4x+ 4- 4+ 3=(x^2+ 4x+ 4)- 1= (x- 2)^2- 1$.

$$latex$$